ML Aggarwal Class 10 Maths Solutions Chapter 22 Probability

Question 1. A box contains 600 screws, one-tenth are rusted. One screw is taken out at random from this box. Find the probability that it is a good screw.
Answer: Out of 600 screws, one-tenth (which equals 60) are rusted. This leaves 540 screws in good working condition. If we define E1 as the event of drawing a good screw, then there are 540 favorable outcomes out of a total of 600 possible outcomes. Using the probability formula: P(E1) = 540/600 = 9/10.
In simple words: Nine out of every ten screws are good. If you pick one at random, the chance it will be good is 9/10.

Exam Tip: Always simplify fractions to their lowest terms - 540/600 reduces to 9/10. Make sure your favorable and total outcomes are clearly labeled and counted correctly.

Question 2. In a lottery, there are 5 prized tickets and 995 blank tickets. A person buys a lottery ticket. Find the probability of his winning a prize.
Answer: The lottery contains 995 blank tickets and 5 winning tickets, making the total 1000 tickets. If E1 represents the event of drawing a prized ticket, then there are 5 favorable outcomes among 1000 total possible outcomes. The probability formula gives: P(E1) = 5/1000 = 1/200.
In simple words: Only 5 tickets out of 1000 can win. Your chance of getting a prize ticket is 1 in 200.

Exam Tip: Don't forget to reduce the fraction - 5/1000 simplifies to 1/200. The winning tickets are the favorable outcomes; all tickets (both winning and blank) form the total possible outcomes.

Question 3. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Answer: The lot contains 12 defective pens and 132 good pens, giving a total of 144 pens. Let E1 be the event of drawing a good pen, with 132 favorable outcomes. Using the probability formula: P(E1) = 132/144 = 11/12.
In simple words: Out of 144 pens, 132 are good. The probability of picking a good pen is 11/12.

Exam Tip: Always add up the counts correctly to find the total (12 + 132 = 144). When simplifying 132/144, divide both by their GCD (12) to get 11/12.

Question 4. Two players, Sania and Sonali, play a tennis match. It is known that the probability of Sania winning the match is 0.69. What is the probability of Sonali winning?
Answer: In any match, either Sania wins or she does not. These are complementary events, so their probabilities must add to 1. Since P(Sania wins) = 0.69, we can find the probability of the opposite outcome: P(Sonali wins) = 1 - 0.69 = 0.31.
In simple words: If Sania has a 0.69 chance of winning, then Sonali's winning probability is 1 - 0.69 = 0.31.

Exam Tip: Remember the complement rule - if an event has probability p, the opposite event has probability 1 - p. These two probabilities always add to 1.

Question 5. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?
Answer:
(i) The bag holds 3 red balls and 5 black balls, for a total of 8 balls. Let E1 be the event of drawing a red ball, giving 3 favorable outcomes. The probability is: P(E1) = 3/8.
(ii) Not drawing a red ball means drawing a black ball instead. With 5 black balls out of 8 total, the probability is: P(E2) = 5/8.
In simple words: Three out of eight balls are red, so the chance of red is 3/8. Five out of eight are black, so not red is 5/8. Notice these add to 1.

Exam Tip: For "not red," recognize it means "black" in this case. The probabilities 3/8 and 5/8 are complements and must sum to 1, confirming your answer.

Question 6. A letter is chosen from the word 'TRIANGLE'. What is the probability that it is a vowel?
Answer: The word TRIANGLE contains 8 letters. The vowels in this word are I, A, and E - that is 3 vowels. If E1 is the event of choosing a vowel, there are 3 favorable outcomes among 8 total letters. Therefore: P(E1) = 3/8.
In simple words: TRIANGLE has 8 letters, and 3 of them are vowels (I, A, E). The chance of picking a vowel is 3/8.

Exam Tip: Count the total letters carefully, including repeated letters if any. Identify all vowels (A, E, I, O, U) present in the word before calculating the probability.

Question 7. A letter of English alphabet is chosen at random. Determine probability that the letter is consonant.
Answer: The English alphabet has 26 letters in total. Of these, 5 are vowels (A, E, I, O, U), leaving 21 consonants. Let E1 be the event of choosing a consonant, with 21 favorable outcomes. The probability is: P(E1) = 21/26.
In simple words: There are 26 letters in English. 21 of them are consonants. So the chance of picking a consonant is 21/26.

Exam Tip: Always remember there are 5 vowels in English (A, E, I, O, U), making 26 - 5 = 21 consonants. The fraction 21/26 cannot be reduced further.

Question 8. A bag contains 5 white, 2 red and 3 black balls. A ball is drawn at random. What is the probability that the ball drawn is red ball?
Answer: The total number of balls in the bag is 5 + 2 + 3 = 10. There are exactly 2 red balls. If we consider drawing a red ball as the favorable outcome, then: P(red) = 2/10 = 1/5.
In simple words: Out of 10 balls, 2 are red. The probability of drawing a red ball is 1/5.

Exam Tip: Count all balls to find the total (5 + 2 + 3 = 10). Always reduce the final fraction - 2/10 simplifies to 1/5 by dividing both by 2.

Question 9. A box contains 7 blue, 8 white and 5 black marbles. If a marble is drawn at random from the box, what is the probability that it will be (i) black? (ii) blue or black? (iii) not black? (iv) green?
Answer:
(i) The total number of marbles is 7 + 8 + 5 = 20. With 5 black marbles, the probability of drawing a black marble is: P(E1) = 5/20 = 1/4.
(ii) Marbles that are either blue or black total 7 + 5 = 12. The probability is: P(E2) = 12/20 = 3/5.
(iii) "Not black" includes blue and white marbles: 7 + 8 = 15 marbles. The probability is: P(E3) = 15/20 = 3/4.
(iv) There are no green marbles in the box, so the number of favorable outcomes is 0. The probability is: P(E4) = 0/20 = 0.
In simple words: Black marbles make up 1/4 of the total. Blue or black together are 3/5. Not black (blue and white) is 3/4. Green marbles don't exist, so that probability is 0.

Exam Tip: For compound events like "blue or black," add the counts of both colors. Remember that an impossible event has probability 0. Always check that your fractions reduce completely.

Question 10. A bag contains 6 red balls, 8 white balls, 5 green balls and 3 black balls. One ball is drawn at random from the bag. Find the probability that the ball is : (i) white (ii) red or black (iii) not green (iv) neither white nor black.
Answer:
(i) The total number of balls is 6 + 8 + 5 + 3 = 22. There are 8 white balls, so: P(E1) = 8/22 = 4/11.
(ii) Red and black balls together: 6 + 3 = 9. The probability is: P(E2) = 9/22.
(iii) "Not green" means any ball except green: 6 + 3 + 8 = 17 balls. The probability is: P(E3) = 17/22.
(iv) "Neither white nor black" means either red or green: 6 + 5 = 11 balls. The probability is: P(E4) = 11/22 = 1/2.
In simple words: White balls are 4/11 of all balls. Red or black total 9/22. Not green (any other color) is 17/22. Red and green only is 1/2.

Exam Tip: Be careful with "neither...nor" statements - it excludes both mentioned colors. Always simplify your final answer: 11/22 = 1/2, and 8/22 = 4/11.

Question 11. A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Peter, a trader, will only accept the shirts which are good, but Salim, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that (i) it is acceptable to Peter? (ii) it is acceptable to Salim?
Answer:
(i) Peter accepts only good shirts. There are 88 good shirts out of 100 total. The probability is: P(E1) = 88/100 = 22/25.
(ii) Salim rejects only shirts with major defects, which means he accepts good and minor-defect shirts. These total 88 + 8 = 96 out of 100. The probability is: P(E2) = 96/100 = 24/25.
In simple words: Peter wants only perfect shirts - 22 out of every 25. Salim accepts anything except major defects - 24 out of every 25.

Exam Tip: Understand the traders' criteria carefully. Peter is strict (good only), while Salim tolerates minor defects but not major ones. Simplify by finding the GCD: 88/100 and 96/100 both reduce by dividing by 4.

Question 12. A die is thrown once. What is the probability that the (i) number is even (ii) number is greater than 2?
Answer:
(i) When a die is thrown once, the sample space is {1, 2, 3, 4, 5, 6} with 6 equally likely outcomes. The even numbers are 2, 4, and 6, giving 3 favorable outcomes. The probability is: P(E1) = 3/6 = 1/2.
(ii) Numbers greater than 2 are 3, 4, 5, and 6, which total 4 favorable outcomes. The probability is: P(E2) = 4/6 = 2/3.
In simple words: A die has six faces. Half of them (3 faces) show even numbers, so the chance is 1/2. Four faces show numbers bigger than 2, so that chance is 2/3.

Exam Tip: Always list the sample space first to avoid missing outcomes. For "greater than 2," the number 2 itself is not included - start from 3. Remember to simplify: 3/6 = 1/2 and 4/6 = 2/3.

Question 13. In a single throw of a die, find the probability of getting : (i) an odd number (ii) a number less than 5 (iii) a number greater than 5 (iv) a prime number (v) a number less than 7 (vi) a number divisible by 3 (vii) a number between 3 and 6 (viii) a number divisible by 2 or 3.
Answer:
In a single die throw, the sample space is {1, 2, 3, 4, 5, 6}.
(i) Odd numbers are 1, 3, 5. There are 3 favorable outcomes. P(E) = 3/6 = 1/2.
(ii) Numbers less than 5 are 1, 2, 3, 4. There are 4 favorable outcomes. P(E) = 4/6 = 2/3.
(iii) Only the number 6 is greater than 5. There is 1 favorable outcome. P(E) = 1/6.
(iv) Prime numbers are 2, 3, 5. There are 3 favorable outcomes. P(E) = 3/6 = 1/2.
(v) Numbers less than 7 are all six numbers: 1, 2, 3, 4, 5, 6. There are 6 favorable outcomes. P(E) = 6/6 = 1.
(vi) Numbers divisible by 3 are 3 and 6. There are 2 favorable outcomes. P(E) = 2/6 = 1/3.
(vii) Numbers between 3 and 6 (not including 3 and 6) are 4 and 5. There are 2 favorable outcomes. P(E) = 2/6 = 1/3.
(viii) Numbers divisible by 2 (even) are 2, 4, 6. Numbers divisible by 3 are 3, 6. Combined without repetition: 2, 3, 4, 6. There are 4 favorable outcomes. P(E) = 4/6 = 2/3.
In simple words: Half the faces show odd numbers (1/2). Four out of six are below 5 (2/3). Only one face (6) exceeds 5 (1/6). Three faces show primes (1/2). All faces are below 7 (certainty = 1). Two faces show multiples of 3 (1/3). Two faces are strictly between 3 and 6 (1/3). Four faces are divisible by 2 or 3 (2/3).

Exam Tip: For "between 3 and 6," clarify whether endpoints are included. Here, 4 and 5 are strictly between. For "divisible by 2 or 3," count each number once - the number 6 appears in both categories but is counted only once. Always simplify fractions: 3/6 = 1/2, 4/6 = 2/3, 2/6 = 1/3, 6/6 = 1.

Question 13. When a die is thrown once, find the probability of getting
(i) an odd number
(ii) a number less than 5
(iii) a number greater than 5
(iv) a prime number
(v) a number less than 7
(vi) a number divisible by 3
(vii) a number between 3 and 6
(viii) a number divisible by 2 or 3
Answer:
(i) Let E be the event of getting an odd number, so E = {1, 3, 5}. The count of favourable outcomes is 3.

\( P(E) = \frac{3}{6} = \frac{1}{2} \)

The probability of getting an odd number is \( \frac{1}{2} \).

(ii) Let E₁ be the event of getting a number less than 5, so E₁ = {1, 2, 3, 4}. The count of favourable outcomes is 4.

\( P(E_1) = \frac{4}{6} = \frac{2}{3} \)

The probability of getting a number less than 5 is \( \frac{2}{3} \).

(iii) Let E₂ be the event of getting a number greater than 5, so E₂ = {6}. The count of favourable outcomes is 1.

\( P(E_2) = \frac{1}{6} \)

The probability of getting a number greater than 5 is \( \frac{1}{6} \).

(iv) Let E₃ be the event of getting a prime number, so E₃ = {2, 3, 5}. The count of favourable outcomes is 3.

\( P(E_3) = \frac{3}{6} = \frac{1}{2} \)

The probability of getting a prime number is \( \frac{1}{2} \).

(v) Let E₄ be the event of getting a number less than 7, so E₄ = {1, 2, 3, 4, 5, 6}. The count of favourable outcomes is 6.

\( P(E_4) = \frac{6}{6} = 1 \)

The probability of getting a number less than 7 is 1.

(vi) Let E₅ be the event of getting a number divisible by 3, so E₅ = {3, 6}. The count of favourable outcomes is 2.

\( P(E_5) = \frac{2}{6} = \frac{1}{3} \)

The probability of getting a number divisible by 3 is \( \frac{1}{3} \).

(vii) Let E₆ be the event of getting a number between 3 and 6, so E₆ = {4, 5}. The count of favourable outcomes is 2.

\( P(E_6) = \frac{2}{6} = \frac{1}{3} \)

The probability of getting a number between 3 and 6 is \( \frac{1}{3} \).

(viii) Let E₇ be the event of getting a number divisible by 2 or 3, so E₇ = {2, 3, 4, 6}. The count of favourable outcomes is 4.

\( P(E_7) = \frac{4}{6} = \frac{2}{3} \)

The probability of getting a number divisible by 2 or 3 is \( \frac{2}{3} \).
In simple words: Probability means the chance something will happen. For a single die showing numbers 1 to 6, divide the count of numbers matching your condition by 6 to find the probability.

Exam Tip: Always identify the total number of possible outcomes first, then count how many match the condition asked. The probability formula is always (matching outcomes) ÷ (total outcomes).

Question 14. A die has 6 faces marked by the given numbers as shown below: 1, 2, 3, -1, -2, -3. The die is thrown once. What is the probability of getting
(i) a positive integer
(ii) an integer greater than -3
(iii) the smallest integer
Answer:
When the die is thrown once, Sample space = {1, 2, 3, -1, -2, -3}.

(i) Let E₁ be the event of getting a positive integer, so E₁ = {1, 2, 3}. The count of favourable outcomes is 3.

\( P(E_1) = \frac{3}{6} = \frac{1}{2} \)

The probability of getting a positive integer is \( \frac{1}{2} \).

(ii) Let E₂ be the event of getting an integer greater than -3, so E₂ = {-2, -1, 1, 2, 3}. The count of favourable outcomes is 5.

\( P(E_2) = \frac{5}{6} \)

The probability of getting an integer greater than -3 is \( \frac{5}{6} \).

(iii) Let E₃ be the event of getting the smallest integer, so E₃ = {-3}. The count of favourable outcomes is 1.

\( P(E_3) = \frac{1}{6} \)

The probability of getting the smallest integer is \( \frac{1}{6} \).
In simple words: Look at all six numbers on the die. Count how many match what you are looking for. Divide that count by 6 to get the probability.

Exam Tip: For problems with negative numbers, be careful to identify which numbers are greater than or less than a given value. The smallest number on this die is -3.

Question 15. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (shown in the adjoining figure) and these are equally likely outcomes. What is the probability that it will point at
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?
Answer:
Sample space = {1, 2, 3, 4, 5, 6, 7, 8}.

(i) Let E₁ be the event of getting 8, so E₁ = {8}. The count of favourable outcomes is 1.

\( P(E_1) = \frac{1}{8} \)

The probability of getting 8 is \( \frac{1}{8} \).

(ii) Let E₂ be the event of getting an odd number, so E₂ = {1, 3, 5, 7}. The count of favourable outcomes is 4.

\( P(E_2) = \frac{4}{8} = \frac{1}{2} \)

The probability of getting an odd number is \( \frac{1}{2} \).

(iii) Let E₃ be the event of getting a number greater than 2, so E₃ = {3, 4, 5, 6, 7, 8}. The count of favourable outcomes is 6.

\( P(E_3) = \frac{6}{8} = \frac{3}{4} \)

The probability of getting a number greater than 2 is \( \frac{3}{4} \).

(iv) Let E₄ be the event of getting a number less than 9, so E₄ = {1, 2, 3, 4, 5, 6, 7, 8}. The count of favourable outcomes is 8.

\( P(E_4) = \frac{8}{8} = 1 \)

The probability of getting a number less than 9 is 1.
In simple words: The spinner can land on any of 8 numbers. Count how many numbers satisfy the condition. Divide that count by 8.

Exam Tip: When a probability equals 1, the event is certain to happen - in this case, all numbers on the spinner are less than 9, so it will always occur.

Question 16. Find the probability that the month of January may have 5 Mondays in
(i) a leap year
(ii) a non-leap year
Answer:
January has 31 days. An ordinary year has 365 days, while a leap year has 366 days.

(i) In a leap year's January, there are 31 days, which makes 4 complete weeks plus 3 extra days. To have 5 Mondays, we need a Monday among these remaining 3 days.

The 3 remaining days can be any of these combinations: (Monday, Tuesday, Wednesday), (Tuesday, Wednesday, Thursday), (Wednesday, Thursday, Friday), (Thursday, Friday, Saturday), (Friday, Saturday, Sunday), (Saturday, Sunday, Monday), (Sunday, Monday, Tuesday).

Among these 7 possible combinations, Monday appears in 3 of them.

\( P(\text{having 5 Mondays}) = \frac{3}{7} \)

The probability that January has 5 Mondays in a leap year is \( \frac{3}{7} \).

(ii) In a non-leap year's January, there are 31 days, which makes 4 complete weeks plus 3 extra days. One of these 3 remaining days could be a Monday.

The 3 remaining days can be any of these combinations: (Monday, Tuesday, Wednesday), (Tuesday, Wednesday, Thursday), (Wednesday, Thursday, Friday), (Thursday, Friday, Saturday), (Friday, Saturday, Sunday), (Saturday, Sunday, Monday), (Sunday, Monday, Tuesday).

Among these 7 possible combinations, Monday appears in 3 of them.

\( P(\text{having 5 Mondays}) = \frac{3}{7} \)

The probability that January has 5 Mondays in a non-leap year is \( \frac{3}{7} \).
In simple words: Any month with 31 days breaks into 4 full weeks plus 3 extra days. The chance of getting 5 of the same day depends on which days those 3 extra days are.

Exam Tip: For calendar probability questions, first break the total days into complete weeks plus remainder days. The probability depends only on the remainder, not on whether it is a leap year (both give the same answer here).

Question 17. Find the probability that the month of February may have 5 Wednesdays in
(i) a leap year
(ii) a non-leap year
Answer:
In February, there are 29 days in a leap year and 28 days in a non-leap year.

(i) In a leap year's February, there are 29 days, which equals 4 complete weeks plus 1 extra day. To have 5 Wednesdays, this one remaining day must be a Wednesday.

\( P(\text{having 5 Wednesdays}) = \frac{1}{7} \)

The probability that February has 5 Wednesdays in a leap year is \( \frac{1}{7} \).

(ii) In a non-leap year's February, there are 28 days, which equals exactly 4 complete weeks with no extra days.

\( P(\text{having 5 Wednesdays}) = \frac{0}{7} = 0 \)

The probability that February has 5 Wednesdays in a non-leap year is 0.
In simple words: A leap year February has one extra day after 4 complete weeks, so 5 Wednesdays might happen. A regular February has exactly 4 weeks with no remainder, so 5 Wednesdays cannot happen.

Exam Tip: When a probability is 0, the event is impossible. Notice that February in a non-leap year never has 5 of any day because it has exactly 28 days - a perfect number of weeks.

Question 18. Sixteen cards are labelled as a, b, c, ..., m, n, o, p. They are put in a box and shuffled. A boy is asked to draw a card from the box. What is the probability that the card drawn is
(i) a vowel
(ii) a consonant
(iii) none of the letters of the word median
Answer:
When drawing a card from the box, Sample space = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p}.

(i) Let E₁ be the event of drawing a vowel card, so E₁ = {a, e, i, o}. The count of favourable outcomes is 4.

\( P(E_1) = \frac{4}{16} = \frac{1}{4} \)

The probability of drawing a vowel card is \( \frac{1}{4} \).

(ii) Let E₂ be the event of drawing a consonant card. Since there are 4 vowels among the 16 letters, the number of consonants = 16 - 4 = 12. The count of favourable outcomes is 12.

\( P(E_2) = \frac{12}{16} = \frac{3}{4} \)

The probability of drawing a consonant card is \( \frac{3}{4} \).

(iii) Let E₃ be the event of drawing a card that does not contain letters from the word median. The word "median" has 6 letters: m, e, d, i, a, n. So the number of other letters = 16 - 6 = 10. The count of favourable outcomes is 10.

\( P(E_3) = \frac{10}{16} = \frac{5}{8} \)

The probability of drawing a card with letters not in the word 'median' is \( \frac{5}{8} \).
In simple words: Out of 16 cards, identify how many match what you need. Divide that number by 16. For example, there are 4 vowels, so the vowel probability is 4 divided by 16, which simplifies to 1/4.

Exam Tip: Always count the total possible outcomes first. When working with letters, identify vowels (a, e, i, o, u) and consonants carefully. Simplify the fraction to lowest terms.

Question 19. Each of the letters of the word 'BOUNDARIES' is written on identical cards and put in the bag. They are well-shuffled. If a card is drawn at random, what is the probability that the letter is
(i) a consonant?
(ii) one of the letters of the word 'LUCKNOW'?
(iii) one of the letters of the word 'INDIA'?
Answer:
(i) In the word 'BOUNDARIES', the consonants are B, N, D, R, S - a total of 5 consonants.

Number of favourable outcomes = 5
Total number of letters in 'BOUNDARIES' = 10
Number of possible outcomes = 10

\( P(\text{consonant}) = \frac{5}{10} = \frac{1}{2} \)

The probability of getting a consonant is \( \frac{1}{2} \).

(ii) The different letters in 'LUCKNOW' that also appear in 'BOUNDARIES' are U, N, O - a total of 3 letters.

Number of favourable outcomes = 3

\( P(\text{letter of 'LUCKNOW'}) = \frac{3}{10} \)

The probability of getting one of the letters from 'LUCKNOW' is \( \frac{3}{10} \).

(iii) The different letters in 'INDIA' that also appear in 'BOUNDARIES' are N, D, I, A - a total of 4 letters.

Number of favourable outcomes = 4

\( P(\text{letter of 'INDIA'}) = \frac{4}{10} = \frac{2}{5} \)

The probability of getting one of the letters from 'INDIA' is \( \frac{2}{5} \).
In simple words: Count all the letters in 'BOUNDARIES' (there are 10). Then count how many of them match the condition you need. Divide to find the probability.

Exam Tip: When finding common letters between two words, list the unique letters from each word and check which ones overlap. Do not double-count repeated letters - count each unique letter only once.

Question 20. An integer is chosen between 0 and 100. What is the probability that it is (i) divisible by 7? (ii) not divisible by 7?
Answer: When an integer is picked between 0 and 100, the sample space is {1, 2, 3, 4, ........, 99}.

(i) Let E1 be the event of selecting a number divisible by 7.
E1 = {7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98}.
The count of favourable outcomes for E1 is 14.

\( P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{14}{99} \)

The probability of selecting an integer divisible by 7 is \( \frac{14}{99} \).

(ii) Let E2 be the event of selecting a number that is not divisible by 7.
Since 14 numbers between 0 and 100 are divisible by 7, the count of numbers not divisible by 7 = 99 - 14 = 85.

\( P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{85}{99} \)

The probability of selecting an integer not divisible by 7 is \( \frac{85}{99} \).
In simple words: Divisible by 7 means the number splits evenly by 7. Count how many such numbers exist from 1 to 99, then divide by the total count. For numbers NOT divisible by 7, subtract the first count from the total.

Exam Tip: Always count the sample space carefully - between 0 and 100 means 1 to 99, not 0 to 100. Verify that the two probabilities add up to 1 as a quick check.

Question 21. Cards marked with numbers 1, 2, 3, 4, ......, 20 are well-shuffled and a card is drawn at random. What is the probability that the number on the card is: (i) a prime number (ii) divisible by 3 (iii) a perfect square?
Answer: The cards are marked 1, 2, 3, 4, ......, 20 and a card is drawn at random.
Sample space = {1, 2, 3, 4, ......, 20}, which has 20 equally likely outcomes.

(i) Let E1 be the event of selecting a prime number.
E1 = {2, 3, 5, 7, 11, 13, 17, 19}.
The count of favourable outcomes for E1 is 8.

\( P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{8}{20} = \frac{2}{5} \)

The probability of selecting a prime number is \( \frac{2}{5} \).

(ii) Let E2 be the event of selecting a number divisible by 3.
E2 = {3, 6, 9, 12, 15, 18}.
The count of favourable outcomes for E2 is 6.

\( P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{6}{20} = \frac{3}{10} \)

The probability of selecting a number divisible by 3 is \( \frac{3}{10} \).

(iii) Let E3 be the event of selecting a perfect square.
E3 = {1, 4, 9, 16}.
The count of favourable outcomes for E3 is 4.

\( P(E_3) = \frac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \frac{4}{20} = \frac{1}{5} \)

The probability of selecting a perfect square is \( \frac{1}{5} \).
In simple words: A prime number has no factors other than 1 and itself. Divisible by 3 means 3 goes into it evenly. A perfect square is a number like 1, 4, 9, 16 - made by multiplying a whole number by itself.

Exam Tip: Remember to list all elements of each event carefully - a missed number changes the count and gives wrong probability. Always simplify the final fraction to lowest terms.

Question 22. There are 25 discs numbered 1 to 25. They are put in a closed box and shaken thoroughly. A disc is drawn at random from the box. Find the probability that the number on the disc is: (i) an odd number (ii) divisible by 2 and 3 both (iii) a number less than 16.
Answer:
(i) Let E1 be the event of selecting an odd number disc.
E1 = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25}.
The count of favourable outcomes for E1 is 13.

\( P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{13}{25} \)

The probability of selecting an odd number disc is \( \frac{13}{25} \).

(ii) Let E2 be the event of selecting a disc whose number is divisible by both 2 and 3.
E2 = {6, 12, 18, 24}.
The count of favourable outcomes for E2 is 4.

\( P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{4}{25} \)

The probability of selecting a disc whose number is divisible by both 2 and 3 is \( \frac{4}{25} \).

(iii) Let E3 be the event of selecting a disc whose number is less than 16.
E3 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}.
The count of favourable outcomes for E3 is 15.

\( P(E_3) = \frac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \frac{15}{25} = \frac{3}{5} \)

The probability of selecting a disc whose number is less than 16 is \( \frac{3}{5} \).
In simple words: Odd numbers don't split evenly by 2 (like 1, 3, 5...). Divisible by both 2 and 3 means the number must be a multiple of 6. "Less than 16" includes all numbers from 1 up to 15.

Exam Tip: For "divisible by both 2 and 3", find the LCM (least common multiple), which is 6. Multiples of 6 up to 25 are: 6, 12, 18, 24 - count these carefully.

Question 23. A box contains 15 cards numbered 1, 2, 3, ...., 15 which are mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is: (i) odd (ii) prime (iii) divisible by 3 (iv) divisible by 3 and 2 both (v) divisible by 3 or 2 (vi) a perfect square number.
Answer:
(i) Let E1 be the event of selecting an odd number card.
E1 = {1, 3, 5, 7, 9, 11, 13, 15}.
The count of favourable outcomes for E1 is 8.

\( P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{8}{15} \)

The probability of selecting an odd number card is \( \frac{8}{15} \).

(ii) Let E2 be the event of selecting a prime number card.
E2 = {2, 3, 5, 7, 11, 13}.
The count of favourable outcomes for E2 is 6.

\( P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{6}{15} = \frac{2}{5} \)

The probability of selecting a prime number card is \( \frac{2}{5} \).

(iii) Let E3 be the event of selecting a card whose number is divisible by 3.
E3 = {3, 6, 9, 12, 15}.
The count of favourable outcomes for E3 is 5.

\( P(E_3) = \frac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \frac{5}{15} = \frac{1}{3} \)

The probability of selecting a card whose number is divisible by 3 is \( \frac{1}{3} \).

(iv) Let E4 be the event of selecting a card whose number is divisible by 3 and 2.
E4 = {6, 12}.
The count of favourable outcomes for E4 is 2.

\( P(E_4) = \frac{\text{No. of favourable outcomes to } E_4}{\text{Total no. of possible outcomes}} = \frac{2}{15} \)

The probability of selecting a card whose number is divisible by 3 and 2 is \( \frac{2}{15} \).

(v) Let E5 be the event of selecting a card whose number is divisible by 3 or 2.
E5 = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15}.
The count of favourable outcomes for E5 is 10.

\( P(E_5) = \frac{\text{No. of favourable outcomes to } E_5}{\text{Total no. of possible outcomes}} = \frac{10}{15} = \frac{2}{3} \)

The probability of selecting a card whose number is divisible by 3 or 2 is \( \frac{2}{3} \).

(vi) Let E6 be the event of selecting a card whose number is a perfect square.
E6 = {1, 4, 9}.
The count of favourable outcomes for E6 is 3.

\( P(E_6) = \frac{\text{No. of favourable outcomes to } E_6}{\text{Total no. of possible outcomes}} = \frac{3}{15} = \frac{1}{5} \)

The probability of selecting a card whose number is a perfect square is \( \frac{1}{5} \).
In simple words: "Divisible by 3 AND 2" means both conditions must be true - use the word "and". "Divisible by 3 OR 2" means at least one condition is true - use the word "or". For "and", find multiples of 6. For "or", count numbers that fit either rule, but don't count twice.

Exam Tip: For "or" questions, watch for overlap - 6 and 12 are divisible by both, so count them only once. Use a Venn diagram approach if needed to avoid double-counting.

Question 24. Cards bearing numbers 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card which is: (i) a prime number (ii) a number divisible by 4 (iii) a number that is a multiple of 6 (iv) an odd number.
Answer: The cards have numbers 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20. So the total count of cards is 10.

(i) Let E1 be the event of selecting a prime number card.
E1 = {2}.
The count of favourable outcomes for E1 is 1.

\( P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{1}{10} \)

The probability of selecting a prime number card is \( \frac{1}{10} \).

(ii) Let E2 be the event of selecting a card whose number is divisible by 4.
E2 = {4, 8, 12, 16, 20}.
The count of favourable outcomes for E2 is 5.

\( P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{5}{10} = \frac{1}{2} \)

The probability of selecting a card whose number is divisible by 4 is \( \frac{1}{2} \).

(iii) Let E3 be the event of selecting a card whose number is a multiple of 6.
E3 = {6, 12, 18}.
The count of favourable outcomes for E3 is 3.

\( P(E_3) = \frac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \frac{3}{10} \)

The probability of selecting a card whose number is a multiple of 6 is \( \frac{3}{10} \).

(iv) Let E4 be the event of selecting a card with an odd number.
E4 = {}.
The count of favourable outcomes for E4 is 0.

\( P(E_4) = \frac{\text{No. of favourable outcomes to } E_4}{\text{Total no. of possible outcomes}} = \frac{0}{10} = 0 \)

The probability of selecting a card with an odd number is 0.
In simple words: All the cards in this bag show even numbers only - 2, 4, 6, 8, and so on. Since there are no odd numbers at all, you can never pick an odd card. That's why the probability is zero.

Exam Tip: Notice that all cards bear even numbers - this is a key fact that makes part (iv) have zero probability. When an event is impossible, its probability is always 0.

Question 25. Cards marked with numbers 13, 14, 15, ...., 60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on card drawn is (i) divisible by 5 (ii) a perfect square number.
Answer: The cards are mixed thoroughly and one is drawn at random, meaning all outcomes have an equal chance of happening.
Sample space = {13, 14, 15, ...., 60}, which has 48 equally likely outcomes.

(i) Let E1 be the event of selecting a card whose number is divisible by 5.
E1 = {15, 20, 25, 30, 35, 40, 45, 50, 55, 60}.
The count of favourable outcomes for E1 is 10.

\( P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{10}{48} = \frac{5}{24} \)

The probability of selecting a card whose number is divisible by 5 is \( \frac{5}{24} \).

(ii) Let E2 be the event of selecting a card with a perfect square number.
E2 = {16, 25, 36, 49}.
The count of favourable outcomes for E2 is 4.

\( P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{4}{48} = \frac{1}{12} \)

The probability of selecting a card with a perfect square number is \( \frac{1}{12} \).
In simple words: Count the total cards from 13 to 60: that's 60 - 13 + 1 = 48 cards. Then list which ones are divisible by 5 or are perfect squares, and find the ratio.

Exam Tip: Always count the sample space correctly - from 13 to 60 inclusive is 48 outcomes, not 47. For perfect squares, recall that 4² = 16, 5² = 25, 6² = 36, 7² = 49 are the only ones in this range.

Question 26. Tickets numbered 3, 5, 7, 9, ...., 29 are placed in a box and mixed thoroughly. One ticket is drawn at random from the box. Find the probability that the number on ticket is (i) a prime number (ii) a number less than 16 (iii) a number divisible by 3.
Answer: The tickets are mixed thoroughly and one is drawn at random, meaning all outcomes have an equal chance of happening.
Sample space = {3, 5, 7, 9, ...., 29}, which has 14 equally likely outcomes.

(i) Let E1 be the event of selecting a ticket with a prime number.
E1 = {3, 5, 7, 11, 13, 17, 19, 23, 29}.
The count of favourable outcomes for E1 is 9.

\( P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{9}{14} \)

The probability of selecting a ticket with a prime number is \( \frac{9}{14} \).

(ii) Let E2 be the event of selecting a ticket with a number less than 16.
E2 = {3, 5, 7, 9, 11, 13, 15}.
The count of favourable outcomes for E2 is 7.

\( P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{7}{14} = \frac{1}{2} \)

The probability of selecting a ticket with a number less than 16 is \( \frac{1}{2} \).

(iii) Let E3 be the event of selecting a ticket with a number divisible by 3.
E3 = {3, 9, 15, 21, 27}.
The count of favourable outcomes for E3 is 5.

\( P(E_3) = \frac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \frac{5}{14} \)

The probability of selecting a ticket with a number divisible by 3 is \( \frac{5}{14} \).
In simple words: This is a list of odd numbers starting at 3: {3, 5, 7, 9, ...., 29}. Count them to find the total (14 tickets). Then identify which ones are prime, less than 16, or divisible by 3.

Exam Tip: The sequence 3, 5, 7, 9, ...., 29 consists of 14 odd numbers total. Note that 9, 15, 21, 27 are odd and divisible by 3, so all are part of the sample space.

Question 27. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5 (iv) a prime number less than 30.
Answer: When a disc is drawn at random, all outcomes have an equal chance of occurring. The sample space is \( \{1, 2, 3, \ldots, 90\} \), giving us 90 equally likely outcomes.

(i) For two-digit numbers: The set is \( E_1 = \{10, 11, 12, 13, \ldots, 90\} \), containing 81 numbers. Therefore, \( P(E_1) = \frac{81}{90} = \frac{9}{10} \).

(ii) For perfect square numbers: The set is \( E_2 = \{1, 4, 9, 16, 25, 36, 49, 64, 81\} \), containing 9 numbers. Therefore, \( P(E_2) = \frac{9}{90} = \frac{1}{10} \).

(iii) For numbers divisible by 5: The set is \( E_3 = \{5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90\} \), containing 18 numbers. Therefore, \( P(E_3) = \frac{18}{90} = \frac{1}{5} \).

(iv) For prime numbers less than 30: The set is \( E_4 = \{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\} \), containing 10 numbers. Therefore, \( P(E_4) = \frac{10}{90} = \frac{1}{9} \).
In simple words: Probability tells us the chance of something happening. To find it, we count how many outcomes match what we want, then divide by the total number of possible outcomes.

Exam Tip: Always identify the sample space first, then carefully count the favorable outcomes for each event. For divisibility and perfect squares, listing all numbers ensures accuracy.

Question 28. A bag contains 15 balls of which some are white and others are red. If the probability of drawing a red ball is twice that of a white ball, find the number of white balls in the bag.
Answer: Let the number of white balls be \( x \). Then the number of red balls is \( 15 - x \).

The probability of drawing a white ball is \( P(\text{white}) = \frac{x}{15} \).

The probability of drawing a red ball is \( P(\text{red}) = \frac{15 - x}{15} \).

We are told that \( P(\text{red}) = 2 \times P(\text{white}) \). Setting up this equation:

\( \frac{15 - x}{15} = 2 \times \frac{x}{15} \)

\( \implies 15 - x = 2x \)

\( \implies 3x = 15 \)

\( \implies x = 5 \)

Therefore, there are 5 white balls in the bag.
In simple words: If red balls are twice as likely to be drawn as white balls, then there must be twice as many red balls as white balls. With 15 total balls, this means 5 white and 10 red.

Exam Tip: Use the given probability relationship to set up an equation. Always verify your answer by checking that the calculated probabilities satisfy the given condition.

Question 29. A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball is twice that of a red ball, find the number of balls in the bag.
Answer: Let the number of blue balls be \( x \). Then the total number of balls is \( x + 6 \).

The probability of drawing a blue ball is \( P(\text{blue}) = \frac{x}{x + 6} \).

The probability of drawing a red ball is \( P(\text{red}) = \frac{6}{x + 6} \).

Given that \( P(\text{blue}) = 2 \times P(\text{red}) \):

\( \frac{x}{x + 6} = 2 \times \frac{6}{x + 6} \)

\( \implies \frac{x}{x + 6} = \frac{12}{x + 6} \)

\( \implies x = 12 \)

Therefore, the total number of balls is \( 12 + 6 = 18 \).
In simple words: When one type of ball is twice as likely to be drawn, there must be twice as many of that type. Since 6 red balls exist, we need 12 blue balls, making 18 total.

Exam Tip: Set up the probability relationship carefully. Remember to find the total number of balls, not just the number of blue balls.

Question 30. A bag contains 24 balls of which x are red, 2x are white and 3x are blue. A ball is selected at random. Find the probability that it is (i) white (ii) not red.
Answer: The total number of balls is \( x + 2x + 3x = 24 \).

This gives us \( 6x = 24 \), so \( x = 4 \).

Therefore: Red balls = 4, White balls = 8, Blue balls = 12.

(i) The probability of drawing a white ball is \( P(\text{white}) = \frac{8}{24} = \frac{1}{3} \).

(ii) The total number of non-red balls is \( 8 + 12 = 20 \). The probability of drawing a non-red ball is \( P(\text{not red}) = \frac{20}{24} = \frac{5}{6} \).
In simple words: First, find the value of x using the total. Then calculate each probability by dividing the favorable outcomes by the total number of balls.

Exam Tip: Always find the value of the variable first before calculating any probabilities. For complementary events like "not red," add all other color counts.

Question 31. A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting: (i) '2' of spades (ii) a jack (iii) a king of red colour (iv) a card of diamond (v) a king or a queen (vi) a non-face card (vii) a black face card (viii) a black card (ix) a non-ace (x) non-face card of black colour (xi) neither a spade nor a jack (xii) neither a heart nor a red king.
Answer: Well-shuffling ensures that all outcomes are equally likely. The pack contains 52 cards in total.

(i) There is only one 2 of spades. \( P(\text{2 of spades}) = \frac{1}{52} \).

(ii) There are 4 jacks (one in each suit). \( P(\text{jack}) = \frac{4}{52} = \frac{1}{13} \).

(iii) There are 2 red kings (hearts and diamonds). \( P(\text{red king}) = \frac{2}{52} = \frac{1}{26} \).

(iv) There are 13 diamond cards. \( P(\text{diamond}) = \frac{13}{52} = \frac{1}{4} \).

(v) There are 8 kings and queens (2 in each suit). \( P(\text{king or queen}) = \frac{8}{52} = \frac{2}{13} \).

(vi) There are 12 face cards, so 40 non-face cards. \( P(\text{non-face}) = \frac{40}{52} = \frac{10}{13} \).

(vii) There are 6 black face cards (3 in spades, 3 in clubs). \( P(\text{black face card}) = \frac{6}{52} = \frac{3}{26} \).

(viii) There are 26 black cards. \( P(\text{black}) = \frac{26}{52} = \frac{1}{2} \).

(ix) There are 4 aces, so 48 non-ace cards. \( P(\text{non-ace}) = \frac{48}{52} = \frac{12}{13} \).

(x) There are 3 face cards in each suit. In black suits there are 6 face cards, leaving 20 non-face black cards. \( P(\text{non-face black}) = \frac{20}{52} = \frac{5}{13} \).

(xi) There are 13 spades and 3 other jacks (excluding the spade jack), totaling 16 cards that are either spades or jacks. Cards that are neither: 52 - 16 = 36. \( P(\text{neither spade nor jack}) = \frac{36}{52} = \frac{9}{13} \).

(xii) There are 13 hearts and 1 red king (of diamonds, since the red king of hearts is already in the heart count), totaling 14. Cards that are neither: 52 - 14 = 38. \( P(\text{neither heart nor red king}) = \frac{38}{52} = \frac{19}{26} \).
In simple words: For each event, count the cards that match the condition, then divide by 52. Use the structure of a deck (4 suits, 13 cards per suit, 3 face cards per suit) to help count accurately.

Exam Tip: Memorize the deck structure: 4 suits, 13 cards each, 3 face cards per suit (Jack, Queen, King), 4 of each rank. Draw a quick diagram if needed to avoid counting errors.

Question 32. All the three face cards of spades are removed from a well-shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting (i) a black face card (ii) a queen (iii) a black card (iv) a heart (v) a spade (vi) '9' of black colour.
Answer: After removing 3 face cards from spades, the total number of cards is 52 - 3 = 49.

(i) Since 2 suits are black and each has 3 face cards, but spade face cards are removed, only club face cards remain. Number of black face cards = 3. \( P(\text{black face card}) = \frac{3}{49} \).

(ii) Originally there are 4 queens. Since the queen of spades is removed, 3 queens remain. \( P(\text{queen}) = \frac{3}{49} \).

(iii) Originally there are 26 black cards. Since 3 spade face cards (which are black) are removed, 23 black cards remain. \( P(\text{black card}) = \frac{23}{49} \).

(iv) There are 13 heart cards, and none have been removed. \( P(\text{heart}) = \frac{13}{49} \).

(v) Originally there are 13 spades. Since 3 face cards are removed, 10 spades remain. \( P(\text{spade}) = \frac{10}{49} \).

(vi) There are 2 nines of black colour (9 of spades and 9 of clubs), and neither has been removed. \( P(\text{9 of black colour}) = \frac{2}{49} \).
In simple words: When cards are removed, the total count changes from 52 to 49. Subtract the removed cards from each category, then calculate the new probabilities.

Exam Tip: Always update the total number of cards when cards are removed. Carefully track which cards are no longer in the pack before calculating each probability.

Question 33. From a pack of 52 cards, a black jack, a red queen and two black kings fell down. A card was then drawn from the remaining pack at random. Find the probability that the card drawn is (i) a black card
Answer: Four cards have been removed: 1 black jack, 1 red queen, and 2 black kings. The remaining pack has 52 - 4 = 48 cards.

(i) Originally there are 26 black cards. The removed cards include 1 black jack and 2 black kings, totaling 3 black cards removed. Black cards remaining = 26 - 3 = 23. \( P(\text{black card}) = \frac{23}{48} \).
In simple words: Start with 26 black cards. Remove 3 black cards (1 jack + 2 kings). You have 23 black cards left out of 48 total.

Exam Tip: Keep track of which color each removed card is. Separate the removals by color when adjusting the counts for probability calculations.

Question 33. A black jack, a red queen and two black kings fell down from a pack of playing cards. From the remaining cards, one card is drawn at random. Find the probability of drawing:
(i) a black card
(ii) a king
(iii) a red queen
Answer: Four cards were removed from the deck - one black jack, one red queen, and two black kings. This leaves 48 cards in total (52 - 4 = 48).

(i) A standard deck has 26 black cards overall, split evenly between clubs (13 cards) and spades (13 cards). Since 3 black cards were taken out, the number of black cards remaining is 23 (26 - 3 = 23).

\( P(\text{a black card}) = \frac{23}{48} \)

The probability of drawing a black card equals \( \frac{23}{48} \).

(ii) A deck contains 4 kings in total, but 2 of them were black and are now gone. This means 2 kings remain (4 - 2 = 2).

\( P(\text{a king}) = \frac{2}{48} = \frac{1}{24} \)

The probability of drawing a king equals \( \frac{1}{24} \).

(iii) There are 2 red queens in a standard deck - one from hearts and one from diamonds. Since one red queen was removed, only 1 red queen is left (2 - 1 = 1).

\( P(\text{a red queen}) = \frac{1}{48} \)

The probability of drawing a red queen equals \( \frac{1}{48} \).
In simple words: When some cards are removed from a deck, fewer cards are available to draw. To find the probability, count how many of the cards you want are left, then divide by the total cards remaining.

Exam Tip: Always subtract the removed cards from both the total deck (52) and from the specific type you are looking for - a common mistake is to forget to adjust both numbers.

Question 34. Two coins are tossed once. Find the probability of getting:
(i) 2 heads
(ii) atleast one tail
Answer: When two coins are flipped at the same time, there are 4 equally likely possible outcomes: {HH, HT, TH, TT}. So the total number of possible outcomes is 4.

(i) Let A be the event of getting two heads. The only way to get two heads is the outcome {HH}. So the number of results that favor event A is 1.

\( P(\text{two heads}) = \frac{1}{4} \)

The probability of getting two heads is \( \frac{1}{4} \).

(ii) Let B be the event of getting at least one tail. This includes any outcome with one or more tails: {HT, TT, TH}. So the number of results that favor event B is 3.

\( P(\text{atleast one tail}) = \frac{3}{4} \)

The probability of getting at least one tail is \( \frac{3}{4} \).
In simple words: When you flip two coins, write down all the ways they can land. Then count how many of those ways match what you are looking for, and divide by the total number of ways.

Exam Tip: Always list the complete sample space first - missing one outcome can change your final answer, especially in problems with multiple items.

Question 35. Two different coins are tossed simultaneously. Find the probability of getting:
(i) two tails
(ii) one tail
(iii) no tail
(iv) atmost one tail
Answer: When two coins are flipped at the same time, all four possible outcomes are equally likely: {HH, HT, TH, TT}. The total number of possible outcomes is 4.

(i) Let A represent the event of getting two tails. The favorable outcome is {TT}, so there is 1 result that supports event A.

\( P(A) = \frac{1}{4} \)

The probability of getting two tails is \( \frac{1}{4} \).

(ii) Let B represent the event of getting one tail. The favorable outcomes are {HT, TH}, so there are 2 results that support event B.

\( P(B) = \frac{2}{4} = \frac{1}{2} \)

The probability of getting one tail is \( \frac{1}{2} \).

(iii) Let C represent the event of getting no tail (meaning both heads). The favorable outcome is {HH}, so there is 1 result that supports event C.

\( P(C) = \frac{1}{4} \)

The probability of getting no tail is \( \frac{1}{4} \).

(iv) Let D represent the event of getting at most one tail (zero or one tail). The favorable outcomes are {HH, HT, TH}, so there are 3 results that support event D.

\( P(D) = \frac{3}{4} \)

The probability of getting at most one tail is \( \frac{3}{4} \).
In simple words: The word "at most" means that many or fewer. So "at most one tail" includes zero tails and also one tail - it is anything except both tails.

Exam Tip: Pay careful attention to phrases like "at least" (that many or more) and "at most" (that many or fewer) - they change which outcomes you count as favorable.

Question 36. Two different dice are thrown simultaneously. Find the probability of getting:
(i) a number greater than 3 on each dice
(ii) an odd number on both dice
Answer: When two dice are rolled together, there are 36 equally likely possible outcomes (6 × 6 = 36). The full sample space is:

\[ S = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \} \]

(i) Let A be the event of rolling a number greater than 3 on each die. The numbers greater than 3 are 4, 5, and 6. The favorable outcomes are {(4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}. There are 9 favorable results.

\( P(A) = \frac{9}{36} = \frac{1}{4} \)

The probability of getting a number greater than 3 on each die is \( \frac{1}{4} \).

(ii) Let B be the event of rolling an odd number on both dice. The odd numbers are 1, 3, and 5. The favorable outcomes are {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}. There are 9 favorable results.

\( P(B) = \frac{9}{36} = \frac{1}{4} \)

The probability of getting an odd number on both dice is \( \frac{1}{4} \).
In simple words: When rolling two dice, pair up each result on the first die with each result on the second die. Count the pairs that match what you want, then divide by 36.

Exam Tip: For two-dice problems, always use ordered pairs (a, b) where a is the first die and b is the second. The pair (4, 5) is different from (5, 4) - this is why there are 36 outcomes, not fewer.

Question 37. Two different dice are thrown at the same time. Find the probability of getting:
(i) a doublet
(ii) a sum of 8
(iii) sum divisible by 5
(iv) sum of atleast 11
Answer:
(i) A doublet occurs when both dice show the same number. The favorable outcomes are {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}. There are 6 favorable results.

\( P(\text{doublet}) = \frac{6}{36} = \frac{1}{6} \)

The probability of getting a doublet is \( \frac{1}{6} \).

(ii) For a sum of 8, the favorable outcomes are {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}. There are 5 favorable results.

\( P(\text{sum of 8}) = \frac{5}{36} \)

The probability of getting a sum of 8 is \( \frac{5}{36} \).

(iii) For a sum divisible by 5, the sums must be 5, 10, 15, etc. The only possible sums are 5, 10, and 15 (since the minimum sum is 2 and the maximum is 12). The favorable outcomes are {(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)}. There are 7 favorable results.

\( P(\text{sum divisible by 5}) = \frac{7}{36} \)

The probability of getting a sum divisible by 5 is \( \frac{7}{36} \).

(iv) For a sum of at least 11, the favorable outcomes are {(5, 6), (6, 5), (6, 6)}. There are 3 favorable results.

\( P(\text{sum of atleast 11}) = \frac{3}{36} = \frac{1}{12} \)

The probability of getting a sum of at least 11 is \( \frac{1}{12} \).
In simple words: Add up the two numbers on the dice. Then count how many of the 36 possible pairs give you a sum matching your condition.

Exam Tip: For sum-related problems, list the pairs systematically by going through each possible sum - this prevents missing outcomes and reduces careless errors.

Question 38. The following letters A, D, M, N, O, S, U, Y of the English alphabet are written on separate cards and put in a box. The cards are well shuffled and one card is drawn at random. What is the probability that the card drawn is a letter of the word,
(a) MONDAY?
(b) which does not appear in MONDAY?
(c) which appears both in SUNDAY and MONDAY?
Answer: The cards contain the letters {A, D, M, N, O, S, U, Y}. There are 8 cards in total.

(a) The letters that make up the word MONDAY are {M, O, N, D, A, Y}. These are all present among the cards. There are 6 letters of MONDAY among the cards.

\[ P(\text{letter of MONDAY}) = \frac{6}{8} = \frac{3}{4} \]

The probability is \( \frac{3}{4} \).

(b) The letters that are not in MONDAY are {S, U}. There are 2 letters among the cards that do not appear in MONDAY.

\[ P(\text{not a letter of MONDAY}) = \frac{2}{8} = \frac{1}{4} \]

The probability is \( \frac{1}{4} \).

(c) The letters that appear in both SUNDAY and MONDAY are {N, D, A, Y}. There are 4 such letters among the cards.

\[ P(\text{letter in both SUNDAY and MONDAY}) = \frac{4}{8} = \frac{1}{2} \]

The probability is \( \frac{1}{2} \).
In simple words: Find which letters from the cards match what you are looking for. Divide that count by 8 to get the probability.

Exam Tip: Always check which letters appear in each word carefully - common mistakes happen when you forget that a letter like A appears in both SUNDAY and MONDAY.

Question 39. In a T.V. show, a contestant opts for video call a friend life line to get an answer from three of his friends, named Amar, Akbar and Anthony. The question which he asks from one of his friends has four options. Find the probability that:
(a) Akbar is chosen for the call.
(b) Akbar couldn't give the correct answer.
Answer:
(a) Three friends can be called: Amar, Akbar, and Anthony. Each friend has an equal chance of being selected. So there is 1 favorable outcome (Akbar is chosen) out of 3 total possibilities.

\( P(\text{Akbar is chosen}) = \frac{1}{3} \)

The probability that Akbar is chosen for the call is \( \frac{1}{3} \).

(b) The question has four answer choices, and only one of them is correct. So there are 3 incorrect answers out of 4 total options. The probability that Akbar picks a wrong answer (assuming he guesses randomly) is \( \frac{3}{4} \).

\( P(\text{Akbar gives wrong answer}) = \frac{3}{4} \)

The probability that Akbar couldn't give the correct answer is \( \frac{3}{4} \).
In simple words: In part (a), divide 1 (the number of ways to pick Akbar) by 3 (the total friends). In part (b), divide 3 (the wrong choices) by 4 (all choices).

Exam Tip: These two parts are independent events - the probability of choosing a particular friend does not affect the probability of getting an answer right or wrong.

Question 1. Which of the following cannot be the probability of an event?
(1) 0.7
(2) \( \frac{2}{3} \)
(3) - 1.5
(4) 15%
Answer: (3) - 1.5
In simple words: Probability values must always be between 0 and 1 (including both 0 and 1). A negative number like - 1.5 can never be a probability because it falls outside this range.

Exam Tip: Remember that probability is always non-negative - it can be zero (impossible event), one (certain event), or anywhere in between, but never below zero or above one.

Question 2. If the probability of an event is p, then the probability of its complementary event will be
(1) p - 1
(2) p
(3) 1 - p
(4) \( 1 - \frac{1}{p} \)
Answer: (3) 1 - p
In simple words: If something has a chance p of happening, then it has a chance of 1 - p of not happening. These two chances add up to 1, which makes sense because one of these two things must occur.

Exam Tip: The complementary event is the "opposite" of the original event. Always remember that P(event) + P(complementary event) = 1.

Question 3. Out of one digit prime numbers, one number is selected at random. The probability of selecting an even number is
(1) \( \frac{1}{2} \)
(2) \( \frac{1}{4} \)
(3) \( \frac{4}{9} \)
(4) \( \frac{2}{5} \)
Answer: (2) \( \frac{1}{4} \)
In simple words: The one-digit prime numbers are 2, 3, 5, and 7. Only the number 2 is even. Since 1 out of 4 numbers is even, the probability is \( \frac{1}{4} \).

Exam Tip: A prime number is a number with exactly two factors: 1 and itself. Remember that 1 is not considered prime, and 2 is the only even prime number.

Question 4. When a die is thrown, the probability of getting an odd number less than 3 is
(1) \( \frac{1}{6} \)
(2) \( \frac{1}{3} \)
(3) \( \frac{1}{2} \)
(4) 0
Answer: (1) \( \frac{1}{6} \)
In simple words: When a die is rolled, the outcomes are {1, 2, 3, 4, 5, 6}. The odd numbers are 1, 3, and 5. Of these, only 1 is less than 3. So there is 1 favorable outcome out of 6 possible ones.

Exam Tip: Always be clear about what the question is asking - here it asks for numbers that are both odd AND less than 3, not just one or the other.

Question 5. The probability of getting a number divisible by 3 in throwing a die is
(1) \( \frac{1}{6} \)
(2) \( \frac{1}{3} \)
(3) \( \frac{1}{2} \)
(4) \( \frac{2}{3} \)
Answer: (2) \( \frac{1}{3} \)
In simple words: On a die, the numbers are 1 through 6. Of these, 3 and 6 are divisible by 3. That is 2 numbers out of 6 total, which simplifies to \( \frac{1}{3} \).

Exam Tip: Always simplify fractions in your final answer - \( \frac{2}{6} \) and \( \frac{1}{3} \) are the same probability, but the simplified form is clearer.

Question 6. A fair die is thrown once. The probability of getting an even prime number is
(1) \( \frac{1}{6} \)
(2) \( \frac{2}{3} \)
(3) \( \frac{1}{3} \)
(4) \( \frac{1}{2} \)
Answer: (1) \( \frac{1}{6} \)
In simple words: A prime number has exactly two factors. An even prime number is a prime that is also even - and the only such number is 2. Since there is 1 favorable outcome out of 6 possible outcomes on a die, the probability is \( \frac{1}{6} \).

Exam Tip: Remember that 2 is special - it is the only prime number that is also even. All other primes (3, 5, 7, 11...) are odd.

Question 7. A fair die is thrown once. The probability of getting a composite number is
(1) \( \frac{1}{3} \)
(2) \( \frac{1}{6} \)
(3) \( \frac{2}{3} \)
(4) \( \frac{1}{2} \)
Answer: (1) \( \frac{1}{3} \)
In simple words: A composite number is one that has more than two factors (unlike a prime, which has exactly two factors). On a die, the composite numbers are 4 and 6. Since 2 out of 6 outcomes are composite, the probability is \( \frac{2}{6} = \frac{1}{3} \).

Exam Tip: The numbers 1 through 6 on a die break down as: 1 (neither prime nor composite), 2 and 3 and 5 (prime), and 4 and 6 (composite). Know these classifications to avoid mix-ups.

Question 7. When a die is thrown the sample space is, S = {1, 2, 3, 4, 5, 6}. Let A be the event of getting a composite number, ∴ A = {4, 6}. Hence, the no. of favourable outcomes to A = 2. What is the probability?
Answer: The number of possible outcomes when rolling a die is 6. Among these, the composite numbers are 4 and 6, which gives us 2 favorable outcomes. Using the probability formula, we divide the number of favorable outcomes by the total number of outcomes: P(A) = 2 ÷ 6 = 1 ÷ 3.
In simple words: A composite number is one that has more than two factors. Out of six numbers on a die, two of them (4 and 6) are composite, so the chance is 1 out of 3.

Exam Tip: Always identify which numbers satisfy the given condition before counting favorable outcomes. Remember that 1 is neither prime nor composite.

Question 8. If a fair die is rolled once, then the probability of getting an even number or a number greater than 4 is
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{1}{6} \)
(d) \( \frac{2}{3} \)
Answer: (d) \( \frac{2}{3} \)
In simple words: When you roll a die, the even numbers are 2, 4, and 6. Numbers greater than 4 are 5 and 6. Together, we have 2, 4, 5, and 6 - that is four numbers out of six total. So the probability is 4 out of 6, which simplifies to 2 out of 3.

Exam Tip: When combining conditions with "or", be careful not to count the same outcome twice. The number 6 appears in both conditions, so count it only once.

Question 9. If a letter is chosen at random from the letters of English alphabet, then the probability that it is a letter of the word 'DELHI' is
(a) \( \frac{1}{5} \)
(b) \( \frac{1}{26} \)
(c) \( \frac{5}{26} \)
(d) \( \frac{21}{26} \)
Answer: (c) \( \frac{5}{26} \)
In simple words: The English alphabet has 26 letters total. The word 'DELHI' contains 5 different letters: D, E, L, H, and I. So the chance of picking one of these five letters is 5 out of 26.

Exam Tip: Identify all distinct letters in the given word carefully. Do not count repeated letters more than once if they appear in the word.

Question 10. A card is selected at random from a pack of 52 cards. The probability of its being a red face card is
(a) \( \frac{3}{26} \)
(b) \( \frac{3}{13} \)
(c) \( \frac{2}{13} \)
(d) \( \frac{1}{2} \)
Answer: (a) \( \frac{3}{26} \)
In simple words: Face cards are Jacks, Queens, and Kings. In a pack, there are 3 red face cards from hearts and 3 red face cards from diamonds, making 6 red face cards total. The probability is 6 out of 52, which reduces to 3 out of 26.

Exam Tip: Remember that a standard playing card deck has 52 cards with face cards appearing in all four suits. Red suits are hearts and diamonds.

Question 11. If a card is drawn from a well-shuffled pack of 52 playing cards, then the probability of this card being a king or jack is
(a) \( \frac{1}{26} \)
(b) \( \frac{1}{13} \)
(c) \( \frac{2}{13} \)
(d) \( \frac{4}{13} \)
Answer: (c) \( \frac{2}{13} \)
In simple words: A deck has 4 kings and 4 jacks, giving us 8 cards that are either a king or a jack. Dividing 8 by 52 and simplifying, we get 2 out of 13.

Exam Tip: Well-shuffling ensures all cards are equally likely to be drawn. Count the total favorable cards carefully when multiple conditions apply.

Question 12. The probability that a non-leap year selected at random has 53 Sundays is
(a) \( \frac{1}{365} \)
(b) \( \frac{2}{365} \)
(c) \( \frac{2}{7} \)
(d) \( \frac{1}{7} \)
Answer: (d) \( \frac{1}{7} \)
In simple words: A non-leap year has 365 days, which makes exactly 52 weeks plus 1 extra day. To have 53 Sundays, that one extra day must be a Sunday. Since any day of the week is equally likely to be that extra day, the probability is 1 out of 7.

Exam Tip: Understand the structure of a year in terms of weeks. The single leftover day determines whether there will be 53 of any particular weekday.

Question 13. A bag contains 3 red balls, 5 white balls and 7 black balls. The probability that a ball drawn from the bag at random will be neither red nor black is
(a) \( \frac{1}{5} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{7}{15} \)
(d) \( \frac{8}{15} \)
Answer: (b) \( \frac{1}{3} \)
In simple words: The bag has 3 + 5 + 7 = 15 balls in total. Balls that are neither red nor black must be white. There are 5 white balls, so the probability is 5 out of 15, which simplifies to 1 out of 3.

Exam Tip: When asked for the probability of an event that is "neither A nor B", identify what remains - in this case, only the white balls qualify.

Question 14. A bag contains 4 red balls and 5 green balls. One ball is drawn at random from the bag. The probability of getting either a red ball or a green ball is
(a) \( \frac{4}{9} \)
(b) \( \frac{5}{9} \)
(c) 0
(d) 1
Answer: (d) 1
In simple words: The bag contains only red and green balls. Any ball drawn from the bag must be either red or green. Since one of these two colors is certain to be drawn, the probability is 1 - it is a sure event.

Exam Tip: When every possible outcome satisfies the given condition, the probability equals 1. This represents a certain event.

Question 15. One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is
(a) \( \frac{1}{5} \)
(b) \( \frac{3}{5} \)
(c) \( \frac{4}{5} \)
(d) \( \frac{1}{3} \)
Answer: (a) \( \frac{1}{5} \)
In simple words: The numbers that are multiples of 5 from 1 to 40 are: 5, 10, 15, 20, 25, 30, 35, and 40. That is 8 numbers out of 40 total. The probability is 8 out of 40, which simplifies to 1 out of 5.

Exam Tip: To find multiples of a number within a range, use the formula: total multiples = (largest multiple - smallest multiple) ÷ divisor + 1.

Question 16. If a number is randomly chosen from the numbers 1, 2, 3, 4, ......., 25, then probability of the number to be prime is
(a) \( \frac{7}{25} \)
(b) \( \frac{9}{25} \)
(c) \( \frac{11}{25} \)
(d) \( \frac{13}{25} \)
Answer: (b) \( \frac{9}{25} \)
In simple words: Prime numbers from 1 to 25 are: 2, 3, 5, 7, 11, 13, 17, 19, and 23. That gives us 9 prime numbers. The probability is 9 out of 25.

Exam Tip: Remember that 1 is not considered a prime number. A prime number has exactly two factors: 1 and itself.

Question 17. A box contains 90 cards numbered 1 to 90. If one card is drawn from the box at random, then the probability that the number on the card is a perfect square is
(a) \( \frac{1}{10} \)
(b) \( \frac{9}{100} \)
(c) \( \frac{1}{9} \)
(d) \( \frac{3}{100} \)
Answer: (a) \( \frac{1}{10} \)
In simple words: Perfect squares from 1 to 90 are: 1, 4, 9, 16, 25, 36, 49, 64, and 81. That is 9 perfect squares out of 90 cards total. The probability is 9 out of 90, which reduces to 1 out of 10.

Exam Tip: Always list out the favorable outcomes systematically. For perfect squares, check which whole number squared gives a result within your range.

Question 18. If a (fair) coin is tossed twice, then the probability of getting two heads is
(a) \( \frac{1}{4} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{3}{4} \)
(d) 0
Answer: (a) \( \frac{1}{4} \)
In simple words: When tossing a coin twice, the possible outcomes are: HH, HT, TH, and TT. Only one of these - HH - gives us two heads. So the probability is 1 out of 4.

Exam Tip: List all possible outcomes systematically when dealing with multiple coin tosses. With n tosses, there are 2^n possible outcomes.

Question 19. If two coins are tossed simultaneously, then the probability of getting atleast one head is
(a) \( \frac{1}{4} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{3}{4} \)
(d) 1
Answer: (c) \( \frac{3}{4} \)
In simple words: When two coins are tossed, the sample space is HH, HT, TH, and TT. The outcomes with at least one head are HH, HT, and TH - that is three outcomes out of four. The probability is 3 out of 4.

Exam Tip: "At least one" means one or more. Count all outcomes satisfying this condition, then divide by the total number of possible outcomes.

Question 20. Lakshmi tosses two coins simultaneously. The probability that she gets atmost one head is
(a) 1
(b) \( \frac{3}{4} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{1}{7} \)
Answer: (b) \( \frac{3}{4} \)
In simple words: When two coins are tossed, the possible outcomes are HH, HT, TH, and TT. Outcomes with at most one head are TT, HT, and TH - that is three outcomes. The probability is 3 out of 4.

Exam Tip: "At most one" includes zero heads (TT) and exactly one head (HT, TH). Do not include the outcome with two heads (HH).

Question 21. The probability of getting a bad egg in a lot of 400 eggs is 0.035. The number of bad eggs in the lot is
(a) 7
(b) 14
(c) 21
(d) 28
Answer: (b) 14
In simple words: If the probability of finding a bad egg is 0.035, multiply this by the total number of eggs (400) to find how many bad eggs are expected. Calculating: 0.035 × 400 = 14.

Exam Tip: Use the relationship: Probability = Favorable Outcomes ÷ Total Outcomes. Rearranging, Favorable Outcomes = Probability × Total Outcomes.

Question 22. Event A: The sun will rise from east tomorrow. Event B: It will rain on Monday. Event C: February month has 29 days in a leap year. Which of the above event(s) has probability equal to 1?
(a) all events A, B and C
(b) both events A and B
(c) both events B and C
(d) both events A and C
Answer: (d) both events A and C
In simple words: Events with probability 1 are certain events - they will definitely happen. The sun always rises from the east, and a leap year always has 29 days in February. However, rain on Monday is uncertain, so event B does not have probability 1.

Exam Tip: Probability equals 1 for sure/certain events, 0 for impossible events, and values between 0 and 1 for uncertain events.

Question 23. A bag contains 3 red and 2 blue marbles. A marble is drawn at random. The probability of drawing a black marble is:
(a) 0
(b) \( \frac{1}{5} \)
(c) \( \frac{2}{5} \)
(d) \( \frac{3}{5} \)
Answer: (a) 0
In simple words: The bag contains only red and blue marbles - there are no black marbles in it. Since it is impossible to draw a black marble, the probability is 0.

Exam Tip: When an event cannot possibly occur, its probability is 0. This is called an impossible event.

Assertion Reason Type Questions

Question. Assertion (A): The probability that a leap year has 53 Sundays is \( \frac{2}{7} \). Reason (R): The probability that a non-leap year has 53 Sundays is \( \frac{1}{7} \).
(a) Assertion (A) is true, but Reason (R) is false.
(b) Assertion (A) is false, but Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
In simple words: A leap year has 366 days, which equals 52 weeks plus 2 extra days. These 2 days can form 7 different pairs of consecutive weekdays. Two of these pairs include Sunday: (Saturday, Sunday) and (Sunday, Monday). So P(53 Sundays in leap year) = 2/7. A non-leap year has 365 days (52 weeks plus 1 day), and for 53 Sundays, that single extra day must be Sunday, giving P(53 Sundays in non-leap year) = 1/7. Both statements are true, but they apply to different situations - one has 2 extra days, the other has 1.

Exam Tip: In Assertion-Reason questions, both statements can be true but still not support each other. Check carefully whether the reason actually explains why the assertion is true.

Question. Assertion (A): In a leap year, getting Sunday as one of the extra days has a probability of 2/7.
Reason (R): A leap year contains 366 days.
Answer: A leap year has 366 days, which equals 52 weeks and 2 days. These 2 extra days can fall on any pair of consecutive days in the week. Out of the 7 possible pairs, exactly 2 pairs include Sunday (Saturday-Sunday and Sunday-Monday). Thus, the probability of getting Sunday as one of the extra days is 2/7. Both the Assertion and Reason are correct, and the Reason correctly explains why the Assertion is true.
In simple words: In a leap year, two extra days beyond the 52 complete weeks will include Sunday in exactly 2 out of 7 possible day combinations.

Exam Tip: Remember that a leap year has 366 days = 52 weeks + 2 days; count all possible day pairs including Sunday to verify the probability.

Question. Assertion (A): Two players Sania and Ashma play a tennis match. If the probability of Sania winning the match is 0.79, then the probability of Ashma winning the match is 0.21.
Reason (R): The sum of probabilities of two complementary events is 1.
Answer: Since Sania and Ashma are the only two possible winners, their winning probabilities are complementary events. By the rule that complementary event probabilities sum to 1: P(Sania wins) + P(Ashma wins) = 1. Substituting the given value: 0.79 + P(Ashma wins) = 1, so P(Ashma wins) = 1 - 0.79 = 0.21. Both statements are correct, and the Reason directly supports the Assertion.
In simple words: When two people play a match, the chance one wins and the chance the other wins must add up to 1. If Sania's chance is 0.79, Ashma's must be 0.21.

Exam Tip: Complementary events always sum to probability 1; use this to find the other player's winning probability instantly.

Question. Assertion (A): If the probability of occurrence of an event E is \( \frac{5}{11} \), then the probability of non-occurrence of the event E is \( \frac{7}{11} \).
Reason (R): If E is an event, then \( P(E) + P(\overline{E}) = 1 \).
Answer: The non-occurrence of event E (denoted \( \overline{E} \)) is the complementary event. By the complementary event rule, \( P(E) + P(\overline{E}) = 1 \). Given \( P(E) = \frac{5}{11} \): \
\( P(\overline{E}) = 1 - \frac{5}{11} = \frac{11 - 5}{11} = \frac{6}{11} \).

The Assertion claims \( P(\overline{E}) = \frac{7}{11} \), which is incorrect. The Reason is true and correctly states the complementary event rule.
In simple words: When an event has probability 5/11, the opposite event has probability 1 - 5/11 = 6/11, not 7/11.

Exam Tip: Always check arithmetic when subtracting from 1; a simple calculation error can make an Assertion false even when the rule is correct.

Question. Assertion (A): In a throw of two fair coins once, the probability of getting one head is \( \frac{1}{2} \).
Reason (R): In a throw of two fair coins, the sample space is {HH, HT, TH, TT}.
Answer: When two coins are tossed, there are 4 equally likely outcomes: {HH, HT, TH, TT}. The Reason is correct. Exactly one head appears in two outcomes: HT and TH. So the probability is \( \frac{2}{4} = \frac{1}{2} \). The Assertion is also correct, and the Reason (the sample space) directly justifies this result.
In simple words: Out of four equally likely outcomes when flipping two coins, exactly two show one head and one tail.

Exam Tip: List all sample space outcomes explicitly; verify that exactly 2 of 4 cases match "one head" before calculating the probability.

Question. Assertion (A): The probability of getting a prime number, when a die is thrown once, is \( \frac{2}{3} \).
Reason (R): On the faces of a die, prime numbers are 2, 3, 5.
Answer: A standard die shows the numbers 1, 2, 3, 4, 5, and 6. The prime numbers among these are 2, 3, and 5 (the Reason is correct). There are 3 prime numbers out of 6 total outcomes, so the probability is \( \frac{3}{6} = \frac{1}{2} \), not \( \frac{2}{3} \). The Assertion is false, though the Reason correctly identifies the prime numbers on the die.
In simple words: Only 3 faces show prime numbers (2, 3, 5), not 4, so the probability is 1/2, not 2/3.

Exam Tip: Always count the total number of favorable outcomes and divide by 6; confusing the count with a simplified fraction leads to error.

Question. Assertion (A): A die is thrown once and the probability of getting an even number is \( \frac{2}{3} \).
Reason (R): The sample space for even number on a die is {2, 4, 6}.
Answer: The faces of a die are 1, 2, 3, 4, 5, and 6. The even numbers are 2, 4, and 6, so the Reason correctly states the sample space for even outcomes. There are 3 even numbers out of 6 total outcomes, giving a probability of \( \frac{3}{6} = \frac{1}{2} \), not \( \frac{2}{3} \). The Assertion is false despite the Reason being true.
In simple words: Three even numbers on a die out of six total gives a probability of 1/2, not 2/3.

Exam Tip: Do not confuse the count of favorable outcomes with a probability fraction; always reduce to lowest terms and verify the denominator is the total number of outcomes.

Chapter Test

Question 1. In a single throw of die, find the probability of getting
(i) a number greater than 5
(ii) an odd prime number
(iii) a number which is multiple of 3 or 4
Answer: When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, 6, forming the sample space {1, 2, 3, 4, 5, 6} with six equally likely outcomes.

(i) A number greater than 5 is only {6}. There is 1 favorable outcome, so \( P(\text{number} > 5) = \frac{1}{6} \).

(ii) An odd prime number on a die means {3, 5}. There are 2 favorable outcomes, so \( P(\text{odd prime}) = \frac{2}{6} = \frac{1}{3} \).

(iii) A number that is a multiple of 3 or 4 means {3, 4, 6}. There are 3 favorable outcomes, so \( P(\text{multiple of 3 or 4}) = \frac{3}{6} = \frac{1}{2} \).
In simple words: For each part, identify which face numbers satisfy the condition, count them, and divide by 6 to get the probability.

Exam Tip: Clearly list the favorable outcomes for each condition before calculating; be precise in identifying which numbers satisfy "multiple of 3 or 4".

Question 2. A lot consists of 48 mobile phones of which 42 are good, 3 have only minor defects and 3 have major defects. Varnika will buy a phone if it is good but the trader will only buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is
(i) acceptable to Varnika?
(ii) acceptable to the trader?
Answer: Total phones = 48.

(i) Varnika accepts only good phones. Since 42 phones are good, the number of favorable outcomes is 42. Thus, \( P(\text{acceptable to Varnika}) = \frac{42}{48} = \frac{7}{8} \).

(ii) The trader accepts phones with no major defect, which includes both good phones (42) and phones with minor defects (3). So the number of favorable outcomes is 42 + 3 = 45. Thus, \( P(\text{acceptable to trader}) = \frac{45}{48} = \frac{15}{16} \).
In simple words: Varnika only wants good phones, so her probability is 42 out of 48. The trader accepts good ones and minor-defect ones, so his probability is 45 out of 48.

Exam Tip: Carefully read the acceptance criteria for each person; minor defects are acceptable to the trader but not to Varnika.

Question 3. A bag contains 5 red, 8 white and 7 black balls. A ball is drawn from the bag at random. Find the probability that the drawn ball is
(i) red or white
(ii) not black
(iii) neither white nor black
Answer: Total balls in the bag = 5 + 8 + 7 = 20. All balls are equally likely to be drawn, so the sample space has 20 equally likely outcomes.

(i) Red or white balls = 5 + 8 = 13. Thus, \( P(\text{red or white}) = \frac{13}{20} \).

(ii) Not black means drawing a red or white ball (the same as part i). Thus, \( P(\text{not black}) = \frac{13}{20} \).

(iii) Neither white nor black means only a red ball is drawn. There are 5 red balls, so \( P(\text{red}) = \frac{5}{20} = \frac{1}{4} \).
In simple words: Count the balls that match each condition, then divide by the total of 20 balls to find each probability.

Exam Tip: Recognize that "not black" and "red or white" are equivalent; use complementary events to simplify calculations when helpful.

Question 4. A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is:
(i) white or blue
(ii) red or black
(iii) not white
(iv) neither white nor black?
Answer: Total balls in the bag = 5 + 7 + 4 + 2 = 18. All balls are equally likely to be drawn, so the sample space has 18 equally likely outcomes.

(i) White or blue balls = 5 + 2 = 7. Thus, \( P(\text{white or blue}) = \frac{7}{18} \).

(ii) Red or black balls = 7 + 4 = 11. Thus, \( P(\text{red or black}) = \frac{11}{18} \).

(iii) Not white means any ball except white. Number of non-white balls = 18 - 5 = 13. Thus, \( P(\text{not white}) = \frac{13}{18} \).

(iv) Neither white nor black means red or blue. Red or blue balls = 7 + 2 = 9. Thus, \( P(\text{neither white nor black}) = \frac{9}{18} = \frac{1}{2} \).
In simple words: For each question, add the matching ball colors and divide by 18 to get the probability.

Exam Tip: Use the complementary event rule for "not white" (total minus white) to avoid counting errors; always verify that probabilities are in simplest form.

Question 5. A box contains 20 balls bearing numbers 1, 2, 3, 4, ......, 20. A ball is drawn at random from the box. What is the probability that the number on the ball is
(i) an odd number
(ii) divisible by 2 or 3
(iii) prime number
(iv) not divisible by 10?
Answer: A ball drawn at random from the box means all 20 outcomes are equally likely.

(i) Odd numbers from 1 to 20 are {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}, totaling 10. Thus, \( P(\text{odd}) = \frac{10}{20} = \frac{1}{2} \).

(ii) Numbers divisible by 2: {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} = 10 numbers. Numbers divisible by 3: {3, 6, 9, 12, 15, 18} = 6 numbers. Numbers divisible by both 2 and 3 (i.e., by 6): {6, 12, 18} = 3 numbers. By inclusion-exclusion: \( 10 + 6 - 3 = 13 \). Thus, \( P(\text{divisible by 2 or 3}) = \frac{13}{20} \).

(iii) Prime numbers from 1 to 20 are {2, 3, 5, 7, 11, 13, 17, 19}, totaling 8. Thus, \( P(\text{prime}) = \frac{8}{20} = \frac{2}{5} \).

(iv) Numbers divisible by 10 are {10, 20}, totaling 2. Numbers not divisible by 10 = 20 - 2 = 18. Thus, \( P(\text{not divisible by 10}) = \frac{18}{20} = \frac{9}{10} \).
In simple words: Identify which numbers 1 to 20 match each condition, count them, then divide by 20.

Exam Tip: For "divisible by 2 or 3," use inclusion-exclusion principle to avoid double-counting numbers divisible by both; for "prime," carefully list all primes up to 20 without missing 17 and 19.

Question. Let A be the event 'the number on the ball is odd', then A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}. The number of favourable outcomes to the event A = 10. Find P(odd number).
Answer: The probability equals the count of outcomes that match the condition divided by the total count of all possible outcomes. There are 10 odd numbers from the set of 20 balls. Thus, \( P(\text{odd number}) = \frac{10}{20} = \frac{1}{2} \). The likelihood that a drawn ball shows an odd number is \( \frac{1}{2} \).
In simple words: Half of the numbers from 1 to 20 are odd. So the chance of picking an odd number is 1 out of 2.

Exam Tip: Always count the matching outcomes carefully and write the probability as a reduced fraction. Double-check that your count matches the total outcomes stated.

Question. Let B be the event 'the number on the ball is divisible by 2 or 3', then B = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20}. The number of favourable outcomes to the event B = 13. Find P(number is divisible by 2 or 3).
Answer: For events involving "or" conditions, count all numbers matching either criterion without double-counting. Numbers divisible by 2 or 3 in the set total 13. Therefore, \( P(\text{divisible by 2 or 3}) = \frac{13}{20} \). The likelihood that a drawn ball has a number divisible by 2 or 3 equals \( \frac{13}{20} \).
In simple words: There are 13 numbers that are divisible by 2 or 3 out of 20 total numbers. The chance is 13 out of 20.

Exam Tip: For "or" conditions, make sure you count each number only once, even if it fits both criteria (like 6, which is divisible by both 2 and 3).

Question. Let C be the event 'the number on the ball is prime', then C = {2, 3, 5, 7, 11, 13, 17, 19}. The number of favourable outcomes to the event C = 8. Find P(prime number).
Answer: A prime number is a number with only two divisors - one and itself. Among numbers 1 to 20, there are 8 prime numbers. Thus, \( P(\text{prime number}) = \frac{8}{20} = \frac{2}{5} \). The likelihood that a drawn ball carries a prime number is \( \frac{2}{5} \).
In simple words: There are 8 prime numbers among 1 to 20. The chance of drawing a prime number is 2 out of 5.

Exam Tip: Remember that 1 is NOT a prime number. Count carefully: 2, 3, 5, 7, 11, 13, 17, and 19 are the only primes up to 20.

Question. Let D be the event 'the number on the ball is not divisible by 10', then D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19}. The number of favourable outcomes to the event D = 18. Find P(number is not divisible by 10).
Answer: Numbers divisible by 10 in this range are only 10 and 20. Excluding these two, 18 numbers remain. Hence, \( P(\text{number is not divisible by 10}) = \frac{18}{20} = \frac{9}{10} \). The likelihood that a drawn ball shows a number not divisible by 10 is \( \frac{9}{10} \).
In simple words: Only 10 and 20 are divisible by 10. All other 18 numbers are not. The chance is 9 out of 10.

Exam Tip: For "not" conditions, it may be easier to count the opposite (numbers that ARE divisible) and subtract from the total, rather than listing all excluded numbers.

Question 6. Find the probability that a number selected at random from the numbers 1, 2, 3, ......, 35 is a (i) prime number (ii) multiple of 7 (iii) multiple of 3 or 5
Answer: Selecting a number at random means all outcomes are equally likely.
(i) Prime numbers from 1 to 35 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31. There are 11 such numbers. Thus, \( P(\text{prime number}) = \frac{11}{35} \). The likelihood that the selected number is prime equals \( \frac{11}{35} \).
(ii) Multiples of 7 from 1 to 35 are: 7, 14, 21, 28, 35. There are 5 such numbers. Therefore, \( P(\text{multiple of 7}) = \frac{5}{35} = \frac{1}{7} \). The likelihood that the selected number is a multiple of 7 is \( \frac{1}{7} \).
(iii) Multiples of 3 or 5 from 1 to 35 are: 3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30, 33, 35. There are 16 such numbers. Hence, \( P(\text{multiple of 3 or 5}) = \frac{16}{35} \). The likelihood that the selected number is a multiple of 3 or 5 equals \( \frac{16}{35} \).
In simple words: Count how many numbers match each condition, then divide by 35. Prime numbers are rare, so the probability is smaller. Multiples are more common, so their probability is larger.

Exam Tip: When finding multiples of two different numbers using "or", list all multiples of each, then remove duplicates (like 15, which is a multiple of both 3 and 5) before counting the total.

Question 7. Cards marked with numbers 13, 14, 15, ...., 60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card is (i) divisible by 5 (ii) a number which is a perfect square.
Answer: Selecting a card at random ensures all outcomes are equally likely. The range is from 13 to 60, which gives 60 - 13 + 1 = 48 cards.
(i) Numbers divisible by 5 in this range are: 15, 20, 25, 30, 35, 40, 45, 50, 55, 60. There are 10 such numbers. Therefore, \( P(\text{divisible by 5}) = \frac{10}{48} = \frac{5}{24} \). The likelihood that the number is divisible by 5 equals \( \frac{5}{24} \).
(ii) Perfect squares in this range are: 16, 25, 36, 49. There are 4 such numbers. Thus, \( P(\text{perfect square}) = \frac{4}{48} = \frac{1}{12} \). The likelihood that the number is a perfect square is \( \frac{1}{12} \).
In simple words: Numbers divisible by 5 end in 0 or 5 - count them carefully. Perfect squares are numbers like 4, 9, 16, 25 - only a few exist in the given range.

Exam Tip: When counting from a starting number other than 1, always verify the total count: subtract the start number from the end number and add 1 to get the true total outcomes.

Question 8. A box has cards numbered 14 to 99. Cards are mixed thoroughly and a card is drawn at random from the box. Find the probability that the card drawn from the box has (i) an odd number (ii) a perfect square number.
Answer: Selecting a card at random makes all outcomes equally likely. The range 14 to 99 gives 99 - 14 + 1 = 86 total cards.
(i) Odd numbers from 14 to 99 are: 15, 17, 19, ..., 97, 99. This is an arithmetic sequence with first term 15, last term 99, and common difference 2. The count is (99 - 15)/2 + 1 = 43 odd numbers. Therefore, \( P(\text{odd number}) = \frac{43}{86} = \frac{1}{2} \). The likelihood that an odd number is drawn equals \( \frac{1}{2} \).
(ii) Perfect squares from 14 to 99 are: 16, 25, 36, 49, 64, 81. There are 6 such numbers. Thus, \( P(\text{perfect square number}) = \frac{6}{86} = \frac{3}{43} \). The likelihood that a perfect square number is drawn is \( \frac{3}{43} \).
In simple words: Odd and even numbers are equally spread, so getting an odd is about half. Perfect squares are much rarer, so their chance is very small.

Exam Tip: For a range like 14 to 99, the count is always (end - start + 1). For odd numbers in an even-start range, the first odd is one more than the start.

Question 9. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is four times that of a red ball, find the number of balls in the bag.
Answer: Let the number of blue balls be x, so the total count is x + 5. The probability of drawing a red ball is \( \frac{5}{x+5} \), and the probability of drawing a blue ball is \( \frac{x}{x+5} \). Given that \( P(\text{blue}) = 4 \times P(\text{red}) \), we have \( \frac{x}{x+5} = 4 \times \frac{5}{x+5} \). Simplifying, \( \frac{x}{x+5} = \frac{20}{x+5} \). Therefore, x = 20. The total number of balls = 20 + 5 = 25.
In simple words: If blue balls are four times as likely to be drawn as red balls, there must be four times as many blue balls as red balls. With 5 red balls, there are 20 blue balls, making 25 total.

Exam Tip: When a probability is described as "times" another, set up an equation with the probability ratio and solve for the unknown variable systematically.

Question 10. A bag contains 18 balls out of which x balls are white. (i) If one ball is drawn at random from the bag, what is the probability that it is white ball? (ii) If 2 more white balls are put in the bag, the probability of drawing a white ball will be \( \frac{9}{8} \) times that of probability of white ball coming in part (i). Find the value of x.
Answer: (i) Let A be the event 'a white ball is drawn'. The number of favourable outcomes is x. Therefore, \( P(\text{a white ball is drawn}) = \frac{x}{18} \).
(ii) After adding 2 white balls, the number of white balls becomes x + 2, and the total becomes 20. The new probability is \( P(\text{a white ball is drawn}) = \frac{x+2}{20} \). We are given that \( \frac{x+2}{20} = \frac{9}{8} \times \frac{x}{18} \). Simplifying the right side: \( \frac{x+2}{20} = \frac{9x}{144} = \frac{x}{16} \). Cross-multiplying, \( 16(x+2) = 20x \), which gives \( 16x + 32 = 20x \). Solving, \( 4x = 32 \), so x = 8. The value of x is 8.
In simple words: In part (i), the chance depends on how many white balls there are out of 18 total. In part (ii), adding 2 white balls changes both the count of white balls and the total count. Setting up the ratio between these two probabilities and solving gives us the answer.

Exam Tip: When a problem involves changing the composition of a bag, write the new probability carefully and use the given ratio to set up an equation. Simplify fractions before cross-multiplying to reduce arithmetic errors.

Question 11. A card is drawn from a well shuffled pack of 52 cards. Find the probability that the card drawn is: (i) a red face card (ii) neither a club nor a spade (iii) neither an ace nor a king of red colour (iv) neither a red card nor a queen (v) neither a red card nor a black king.
Answer: Well shuffling ensures all outcomes are equally likely. The total number of outcomes is 52.
(i) Each suit contains one king, one queen, and one jack. There are two red suits (hearts and diamonds). Thus, there are 2 kings, 2 queens, and 2 jacks of red colour, giving 6 red face cards total. Therefore, \( P(\text{a red face card}) = \frac{6}{52} = \frac{3}{26} \). The probability is \( \frac{3}{26} \).
(ii) Clubs and spades together make 26 cards. Cards that are neither clubs nor spades number 52 - 26 = 26 (these are hearts and diamonds). Thus, \( P(\text{neither a club nor spade}) = \frac{26}{52} = \frac{1}{2} \). The probability is \( \frac{1}{2} \).
(iii) There are 2 red kings (one from hearts, one from diamonds) and 4 aces (one in each suit). The count of cards that are neither aces nor red kings is 52 - 4 - 2 = 46. Therefore, \( P(\text{neither an ace nor a king of red colour}) = \frac{46}{52} = \frac{23}{26} \). The probability is \( \frac{23}{26} \).
(iv) There are 26 red cards. There are 4 queens total, but 2 of them are red (already counted in the 26 red cards). The 2 remaining queens are black. Cards that are either red or queens total 26 + 2 = 28. Cards that are neither equal 52 - 28 = 24. Thus, \( P(\text{neither a red card nor queen}) = \frac{24}{52} = \frac{6}{13} \). The probability is \( \frac{6}{13} \).
(v) There are 26 red cards and 2 black kings (one from clubs, one from spades). Cards that are either red or black kings total 26 + 2 = 28. Cards that are neither equal 52 - 28 = 24. Therefore, \( P(\text{neither a red card nor black king}) = \frac{24}{52} = \frac{6}{13} \). The probability is \( \frac{6}{13} \).
In simple words: To find "neither...nor" probabilities, subtract the unwanted cards from 52. Be careful not to double-count cards that meet both conditions (like red queens, which count as both red and queens).

Exam Tip: Always clarify whether a card counted in the first group (like a red queen) should also be counted in the second group. Use a Venn diagram mentally if needed to avoid overlap errors.

Question 12. From pack of 52 playing cards, black jacks, black kings and black aces are removed and then the remaining pack is well-shuffled. A card is drawn at random from the remaining pack. Find the probability of getting (i) a red card (ii) a face card (iii) a diamond or a club (iv) a queen or a spade.
Answer: There are 2 black jacks, 2 black kings, and 2 black aces, totalling 6 cards removed. The remaining pack has 52 - 6 = 46 cards.
(i) There are 26 red cards (13 hearts and 13 diamonds). None of these were removed. Thus, \( P(\text{a red card}) = \frac{26}{46} = \frac{13}{23} \). The probability is \( \frac{13}{23} \).
(ii) Originally, there are 12 face cards. Four were removed (2 black jacks, 2 black kings, but NOT black queens since they are red). The remaining face cards are 12 - 4 = 8. Therefore, \( P(\text{a face card}) = \frac{8}{46} = \frac{4}{23} \). The probability is \( \frac{4}{23} \).
(iii) There are 13 diamond cards (none removed) and 13 club cards. From clubs, one king, one queen, and one ace were removed, leaving 13 - 3 = 10 club cards. Total diamond and club cards = 13 + 10 = 23. Thus, \( P(\text{a diamond or club}) = \frac{23}{46} = \frac{1}{2} \). The probability is \( \frac{1}{2} \).
(iv) There are 13 spade cards (none removed, since only black aces, jacks, and kings were removed from each suit). There are 4 queens originally. The queen of spades was not removed (because only spade aces, jacks, and kings were removed). So, 3 queens remain (the queens of hearts, diamonds, and spades; the queen of clubs is still there, making 4 - 1 = 3 remaining... actually, all 4 queens remain since no queens were removed). Let me reconsider: no queens were removed at all. So all 4 queens remain. However, we need spades and queens without double-counting the queen of spades. The count is 13 spades - 3 removed + 4 queens - 1 (queen of spade already in spades) = 10 + 3 = 13. Thus, \( P(\text{a queen or spade}) = \frac{13}{46} \). The probability is \( \frac{13}{46} \).
In simple words: When cards are removed, first count how many are left, then identify which desired cards remain. For "or" conditions, avoid counting the same card twice.

Exam Tip: Keep careful track of which cards are removed. Note that only black non-face cards (aces) and black face cards (jacks and kings, NOT queens) are taken out in this problem.

Question 13. Two different dice are thrown simultaneously. Find the probability of getting: (i) sum 7 (ii) sum ≤ 3 (iii) sum ≤ 10
Answer: When two different dice are rolled together, the total number of outcomes is 6 × 6 = 36, and all outcomes are equally likely. The sample space is S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}.
(i) Outcomes with sum 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). There are 6 such outcomes. Therefore, \( P(\text{sum of 7}) = \frac{6}{36} = \frac{1}{6} \). The probability is \( \frac{1}{6} \).
(ii) Outcomes with sum ≤ 3 are: (1,1), (1,2), (2,1). There are 3 such outcomes. Thus, \( P(\text{sum} \leq 3) = \frac{3}{36} = \frac{1}{12} \). The probability is \( \frac{1}{12} \).
(iii) Outcomes with sum > 10 are: (5,6), (6,5), (6,6). There are 3 such outcomes. Therefore, outcomes with sum ≤ 10 are 36 - 3 = 33. Thus, \( P(\text{sum} \leq 10) = \frac{33}{36} = \frac{11}{12} \). The probability is \( \frac{11}{12} \).
In simple words: For sum 7, find all ordered pairs that add to 7. For sum ≤ 3, only very small sums (2 and 3) are possible. For sum ≤ 10, it is easier to count the opposite (sum > 10) and subtract from 36.

Exam Tip: When finding outcomes for two dice, remember that (1,6) and (6,1) are different outcomes. For inequalities like ≤ or ≥, it may be quicker to count the opposite and use the complement rule: P(A) = 1 - P(not A).

Question 14. Two dice are thrown together. Find the probability that the product of the numbers on the top of two dice is
(i) 4
(ii) 12
(iii) 7
Answer: When two different dice are rolled, the total number of possible outcomes is 6 × 6, which equals 36, and each outcome has an equal chance of occurring. The sample space for this experiment is:

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

This space contains 36 equally likely outcomes.

(i) Let A be the event where the product of the numbers is 4.

A = {(1, 4), (2, 2), (4, 1)}

\( \therefore P(A) = \frac{3}{36} = \frac{1}{12} \)

So, the probability that the product of the numbers showing on top is 4 equals \( \frac{1}{12} \).

(ii) Let B be the event where the product of the numbers is 12.

B = {(2, 6), (3, 4), (4, 3), (6, 2)}

\( \therefore P(B) = \frac{4}{36} = \frac{1}{9} \)

So, the probability that the product of the numbers showing on top is 12 equals \( \frac{1}{9} \).

(iii) Let C be the event where the product of the numbers is 7.

C = {}

\( \therefore P(C) = \frac{0}{36} = 0 \)

So, the probability that the product of the numbers showing on top is 7 equals 0.
In simple words: To find the probability, count how many pairs of numbers have the product you want, then divide by the total 36 outcomes. For a product of 4, there are 3 pairs - for 12, there are 4 pairs - and for 7, there are no pairs at all since no two dice numbers multiply to give 7.

Exam Tip: Always list the sample space first to make sure you count all 36 outcomes. When finding matching outcomes, check each pair carefully to avoid missing any or counting duplicates. Remember that getting 0 probability means the event is impossible.