Selina Concise Solutions for ICSE Class 8 Mathematics Chapter 23 Probability

Exercise 23

Question 1. A die is thrown, find the probability of getting:
(i) a prime number
(ii) a number greater than 4
(iii) a number not greater than 4.
Answer:
When a standard six-sided die is rolled, the total number of possible outcomes is 6, which are 1, 2, 3, 4, 5, and 6.
(i) The prime numbers on a die are 2, 3, and 5. This gives a total of 3 favourable outcomes.
\( P(\text{getting a prime number}) = \frac{3}{6} = \frac{1}{2} \)
(ii) The numbers greater than 4 on a die are 5 and 6, which gives 2 favourable outcomes.
\( P(\text{getting a number } > 4) = \frac{2}{6} = \frac{1}{3} \)
(iii) The numbers that are not greater than 4 are 1, 2, 3, and 4, which gives 4 favourable outcomes.
\( P(\text{getting a number } \le 4) = \frac{4}{6} = \frac{2}{3} \)
In simple words: A die has 6 sides in total. To find the probability of any of these events, count how many numbers fit the description and divide that count by 6, then reduce the fraction.

Exam Tip: Remember that 1 is neither prime nor composite; the prime numbers on a six-sided die are 2, 3, and 5.

Question 2. A coin is tossed. What is the probability of getting:
(i) a tail? (ii) ahead?
Answer:
On tossing a coin once, there are 2 possible outcomes: a head or a tail.
(i) Favourable outcome of getting a tail = 1
\( P(\text{getting a tail}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{2} \)
(ii) Favourable outcome of getting a head = 1
\( P(\text{getting a head}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{2} \)
In simple words: A coin has only two sides, so the chance of landing on either heads or tails is exactly 1 out of 2.

Exam Tip: Always state the total possible outcomes first to establish the denominator before computing the probability of a specific event.

Question 3. A coin is tossed twice. Find the probability of getting:
(i) exactly one head (ii) exactly one tail
(iii) two tails (iv) two heads
Answer:
When two coins are flipped together, the list of all possible outcomes is HH, HT, TH, and TT. The total number of outcomes is 4.
(i) Favourable outcomes for exactly one head are HT and TH (2 outcomes).
\( P(\text{exactly one head}) = \frac{2}{4} = \frac{1}{2} \)
(ii) Favourable outcomes for exactly one tail are HT and TH (2 outcomes).
\( P(\text{exactly one tail}) = \frac{2}{4} = \frac{1}{2} \)
(iii) Favourable outcome for two tails is TT (1 outcome).
\( P(\text{two tails}) = \frac{1}{4} \)
(iv) Favourable outcome for two heads is HH (1 outcome).
\( P(\text{two heads}) = \frac{1}{4} \)
In simple words: Flipping two coins gives four possible combinations. Count how many of those match your target condition and write that count as a fraction out of four.

Exam Tip: Clearly listing the sample space (HH, HT, TH, TT) at the start of your answer is highly recommended as it helps avoid mistakes when identifying favourable cases.

Question 4. A letter is chosen from the word ‘PENCIL’ what is the probability that the letter chosen is a consonant?
Answer:
The word 'PENCIL' contains a total of 6 letters. Out of these, the consonants are P, N, C, and L, which total 4 favourable outcomes.
\( P(\text{consonant}) = \frac{\text{Total No. of consonants}}{\text{Total No. of letters in the word PENCIL}} = \frac{4}{6} = \frac{2}{3} \)
In simple words: There are four consonants out of six total letters in 'PENCIL'. This simplifies to a two-thirds chance.

Exam Tip: Carefully separate the vowels (E, I) and consonants (P, N, C, L) in the given word before calculating the probability.

Question 5. A bag contains a black ball, a red ball and a green ball, all the balls are identical in shape and size. A ball is drawn from the bag without looking into it. What is the probability that the ball drawn is:
(i) a red ball
(ii) not a red ball
(iii) a white ball.
Answer:
The bag contains one black, one red, and one green ball, making the total number of possible outcomes 3.
(i) There is only 1 red ball in the bag.
\( P(\text{red ball}) = \frac{1}{3} \)
(ii) The balls that are not red are the black ball and the green ball, making 2 favourable outcomes.
\( P(\text{not a red ball}) = \frac{2}{3} \)
(iii) Since there are no white balls in the bag, the number of favourable outcomes is 0.
\( P(\text{white ball}) = \frac{0}{3} = 0 \)
In simple words: Since there are three balls in total, any single ball has a one-third chance of being picked. If a color is not in the bag, its chance is zero.

Exam Tip: An event with a probability of 0 is called an impossible event, which is the case for drawing a white ball here.

Question 6. In a single throw of a die, find the probability of getting a number
(i) greater than 2
(ii) less than or equal to 2
(iii) not greater than 2.
Answer:
A standard six-sided die contains the numbers 1, 2, 3, 4, 5, and 6, giving 6 possible outcomes.
(i) The numbers greater than 2 are 3, 4, 5, and 6, representing 4 favourable outcomes.
\( P(\text{greater than 2}) = \frac{4}{6} = \frac{2}{3} \)
(ii) The numbers that are less than or equal to 2 are 1 and 2, giving 2 favourable outcomes.
\( P(\text{less than or equal to 2}) = \frac{2}{6} = \frac{1}{3} \)
(iii) The numbers not greater than 2 are also 1 and 2, which represents 2 favourable outcomes.
\( P(\text{not greater than 2}) = \frac{2}{6} = \frac{1}{3} \)
In simple words: We find how many numbers on the die fit each condition and divide that number by six. "Less than or equal to 2" and "not greater than 2" mean the exact same numbers.

Exam Tip: Pay close attention to terms like "not greater than 2" as they are equivalent to "less than or equal to 2". Recognizing these phrasing equivalents prevents confusion.

Question 7. A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn is:
(i) a black ball.
(ii) a red ball.
(iii) a white ball.
(iv) not a red ball.
(v) not a black ball.
Answer:
The total number of balls in the bag is the sum of 3 white, 5 black, and 2 red balls, which is:
\( 3 + 5 + 2 = 10 \)
So, there are 10 possible outcomes.
(i) The number of black balls is 5.
\( P(\text{black ball}) = \frac{5}{10} = \frac{1}{2} \)
(ii) The number of red balls is 2.
\( P(\text{red ball}) = \frac{2}{10} = \frac{1}{5} \)
(iii) The number of white balls is 3.
\( P(\text{white ball}) = \frac{3}{10} \)
(iv) The balls that are not red are the white and black balls, which total \( 3 + 5 = 8 \) balls.
\( P(\text{not a red ball}) = \frac{8}{10} = \frac{4}{5} \)
(v) The balls that are not black are the white and red balls, which total \( 3 + 2 = 5 \) balls.
\( P(\text{not a black ball}) = \frac{5}{10} = \frac{1}{2} \)
In simple words: First, add up all the balls to find the total (10). Then, for each part, divide the number of successful choices by 10 and reduce the fraction.

Exam Tip: When finding "not" a certain color, you can either subtract that color's count from the total or add the counts of all other colors.

Question 8. In a single throw of a die, find the probability that the number:
(i) will be an even number.
(ii) will be an odd number.
(iii) will not be an even number.
Answer:
A die has 6 possible outcomes, which are 1, 2, 3, 4, 5, and 6.
(i) The even numbers are 2, 4, and 6, giving 3 favourable outcomes.
\( P(\text{even number}) = \frac{3}{6} = \frac{1}{2} \)
(ii) The odd numbers are 1, 3, and 5, giving 3 favourable outcomes.
\( P(\text{odd number}) = \frac{3}{6} = \frac{1}{2} \)
(iii) Numbers that are not even are odd numbers, which are 1, 3, and 5 (3 outcomes).
\( P(\text{not an even number}) = \frac{3}{6} = \frac{1}{2} \)
In simple words: Exactly half of the numbers on a die are even, and the other half are odd, so any of these options has a one-half chance.

Exam Tip: Since "not even" is exactly the same as "odd", their probabilities are identical. Always explain this relationship in your working.

Question 9. In a single throw of a die, find the probability of getting :
(i) 8
(ii) a number greater than 8
(iii) a number less than 8
Answer:
The outcomes of rolling a die are 1, 2, 3, 4, 5, and 6. The total count of outcomes is 6.
(i) Since 8 is not on a die, the number of favourable outcomes is 0.
\( P(\text{getting 8}) = \frac{0}{6} = 0 \)
(ii) None of the numbers on a die are greater than 8, so there are 0 favourable outcomes.
\( P(\text{number } > 8) = \frac{0}{6} = 0 \)
(iii) All 6 numbers (1, 2, 3, 4, 5, 6) are less than 8, which means there are 6 favourable outcomes.
\( P(\text{number } < 8) = \frac{6}{6} = 1 \)
In simple words: You cannot get an 8 or anything bigger on a regular die, so those chances are zero. Every number is smaller than 8, so that chance is a sure thing, which is 1.

Exam Tip: An event with a probability of 1 is a certain event, while an event with a probability of 0 is an impossible event.

Question 10. Which of the following can not be the probability of an event?
(i) \( \frac{2}{7} \)
(ii) 3.8
(iii) 37%
(iv) -0.8
(v) 0.8
(vi) \( \frac{-2}{5} \)
(vii) \( \frac{7}{8} \)
Answer:
The probability of any event must lie between 0 and 1, inclusive (i.e., \( 0 \le P(E) \le 1 \)). Therefore:
- (ii) 3.8 cannot be a probability because it is greater than 1.
- (iv) -0.8 cannot be a probability because it is less than 0 (negative).
- (vi) \( \frac{-2}{5} \) cannot be a probability because it is less than 0 (negative).
Thus, (ii), (iv), and (vi) cannot be the probability of an event.
In simple words: A probability can never be a negative number, and it can never be greater than 1 (or 100%). This means 3.8, -0.8, and -2/5 are impossible as probabilities.

Exam Tip: Remember that any probability must satisfy the basic inequality \( 0 \le P(E) \le 1 \). No probability can be negative or exceed 100%.

Question 11. A bag contains six identical black balls. A child withdraws one ball from the bag without looking into it. What is the probability that he takes out:
(i) a white ball,
(ii) a black ball
Answer:
There are 6 black balls in the bag, and no other balls. Therefore, the total number of possible outcomes is 6.
(i) Since there are no white balls in the bag, the number of favourable outcomes is 0.
\( P(\text{white ball}) = \frac{0}{6} = 0 \)
(ii) Since all 6 balls in the bag are black, the number of favourable outcomes is 6.
\( P(\text{black ball}) = \frac{6}{6} = 1 \)
In simple words: Since there are only black balls in the bag, you cannot pull out a white ball (chance is 0), and you will definitely pull out a black ball (chance is 1).

Exam Tip: Be careful with situations where all items in a set are identical. Drawing an item of that type is a certain event with a probability of 1.

Question 12. Three identical coins are tossed together. What is the probability of obtaining:
all heads?
exactly two heads?
exactly one head?
no head?
Answer:
When three coins are tossed, the sample space of all possible outcomes consists of 8 elements:
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
So, the total number of outcomes is 8.
- Probability of obtaining all heads:
Favourable outcome is only HHH (1 outcome).
\( P(\text{all heads}) = \frac{1}{8} \)
- Probability of obtaining exactly two heads:
Favourable outcomes are HHT, HTH, and THH (3 outcomes).
\( P(\text{exactly two heads}) = \frac{3}{8} \)
- Probability of obtaining exactly one head:
Favourable outcomes are HTT, THT, and TTH (3 outcomes).
\( P(\text{exactly one head}) = \frac{3}{8} \)
- Probability of obtaining no head:
Favourable outcome is TTT (1 outcome).
\( P(\text{no head}) = \frac{1}{8} \)
In simple words: Tossing three coins gives eight total combinations. Count how many times your specific target happens, then write that as a fraction out of eight.

Exam Tip: Systematically list all 8 outcomes of tossing three coins to ensure you do not miss any cases or miscount favourable outcomes.

Question 13. A book contains 92 pages. A page is chosen at random. What is the probability that the sum of the digits in the page number is 9?
Answer:
The book is numbered from page 1 to 92, meaning there are 92 possible outcomes.
The page numbers where the sum of the digits equals 9 are:
9, 18, 27, 36, 45, 54, 63, 72, 81, and 90.
This gives a total of 10 favourable outcomes.
\( P(\text{sum of digits is 9}) = \frac{10}{92} = \frac{5}{46} \)
In simple words: There are 10 pages between 1 and 92 whose digits add up to 9. Since there are 92 pages in total, the probability is 10 out of 92, which simplifies to 5 out of 46.

Exam Tip: Be thorough when listing numbers to find the digit sum; don't forget the single-digit page '9' and the page '90', which are often overlooked.

Question 14. Two coins are tossed together. What is the probability of getting:
(i) at least one head
(ii) both heads or both tails.
Answer:
When two coins are tossed together, the possible outcomes are HH, HT, TH, and TT. The total number of outcomes is 4.
(i) "At least one head" means we can have 1 or 2 heads. The favourable outcomes are HH, HT, and TH, which is 3 outcomes.
\( P(\text{at least one head}) = \frac{3}{4} \)
(ii) "Both heads or both tails" means HH or TT, giving 2 favourable outcomes.
\( P(\text{both heads or both tails}) = \frac{2}{4} = \frac{1}{2} \)
In simple words: Out of the four possible ways the coins can land, three ways contain at least one head. Two ways have matching sides (either both heads or both tails).

Exam Tip: Note that "at least one head" includes the outcome with two heads (HH) as well as the outcomes with exactly one head (HT, TH).

Question 15. From 10 identical cards, numbered 1, 2, 3, ...... , 10, one card is drawn at random. Find the probability that the number on the card drawn is a multiple of:
(i) 2
(ii) 3
(iii) 2 and 3
(iv) 2 or 3
Answer:
The cards are numbered from 1 to 10, so there are 10 possible outcomes.
(i) Multiples of 2 in this range are 2, 4, 6, 8, and 10 (5 outcomes).
\( P(\text{multiple of 2}) = \frac{5}{10} = \frac{1}{2} \)
(ii) Multiples of 3 in this range are 3, 6, and 9 (3 outcomes).
\( P(\text{multiple of 3}) = \frac{3}{10} \)
(iii) "Multiple of 2 and 3" means a number that is a multiple of both, which is 6 (1 outcome).
\( P(\text{multiple of 2 and 3}) = \frac{1}{10} \)
(iv) "Multiple of 2 or 3" includes any number that is a multiple of 2, a multiple of 3, or both. These are 2, 3, 4, 6, 8, 9, and 10 (7 outcomes).
\( P(\text{multiple of 2 or 3}) = \frac{7}{10} \)
In simple words: We look through the numbers 1 to 10 to see which ones are multiples of 2, 3, both, or either, and then divide how many we find by 10.

Exam Tip: Be careful with "and" versus "or". "And" requires both conditions to be met (common multiples), whereas "or" allows either condition to be met (union of both sets).

Question 16. Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is:
(i) 0
(ii) 12
(iii) less than 12
(iv) less than or equal to 12
Answer:
When two dice are thrown together, there are \( 6 \times 6 = 36 \) possible outcomes. These outcomes range from (1, 1) to (6, 6).
(i) The minimum sum possible is \( 1 + 1 = 2 \). Thus, a sum of 0 is impossible, giving 0 favourable outcomes.
\( P(\text{sum of 0}) = \frac{0}{36} = 0 \)
(ii) The only outcome that gives a sum of 12 is (6, 6), which is 1 favourable outcome.
\( P(\text{sum of 12}) = \frac{1}{36} \)
(iii) All outcomes except (6, 6) have a sum less than 12. This leaves \( 36 - 1 = 35 \) favourable outcomes.
\( P(\text{sum less than 12}) = \frac{35}{36} \)
(iv) Since the maximum sum is 12, all 36 outcomes have a sum less than or equal to 12. This gives 36 favourable outcomes.
\( P(\text{sum } \le 12) = \frac{36}{36} = 1 \)
In simple words: When rolling two dice, the smallest sum you can get is 2 and the largest is 12. So, getting a sum of 0 is impossible, while getting a sum of 12 or less is guaranteed.

Exam Tip: Showing the sample space grid or explaining how the sum bounds (from 2 to 12) work helps justify your answers for impossible and certain events.

Question 17. A die is thrown once. Find the probability of getting:
(i) a prime number
(ii) a number greater than 3
(iii) a number other than 3 and 5
(iv) a number less than 6
(v) a number greater than 6.
Answer:
Throwing a die once gives a total of 6 possible outcomes: 1, 2, 3, 4, 5, and 6.
(i) The prime numbers are 2, 3, and 5 (3 outcomes).
\( P(\text{prime number}) = \frac{3}{6} = \frac{1}{2} \)
(ii) The numbers greater than 3 are 4, 5, and 6 (3 outcomes).
\( P(\text{number } > 3) = \frac{3}{6} = \frac{1}{2} \)
(iii) The numbers other than 3 and 5 are 1, 2, 4, and 6 (4 outcomes).
\( P(\text{number other than 3 and 5}) = \frac{4}{6} = \frac{2}{3} \)
(iv) The numbers less than 6 are 1, 2, 3, 4, and 5 (5 outcomes).
\( P(\text{number less than 6}) = \frac{5}{6} \)
(v) There is no number greater than 6 on a standard die, which means 0 favourable outcomes.
\( P(\text{number } > 6) = \frac{0}{6} = 0 \)
In simple words: Write down all six numbers on a die. Count how many of them fit each of the criteria, and then put that over 6 to find the probability.

Exam Tip: Be sure to exclude 3 and 5 from the total of 6 numbers when answering part (iii), which leaves exactly 4 favourable outcomes.

Question 18. Two coins are tossed together. Find the probability of getting:
(i) exactly one tail
(ii) at least one head
(iii) no head
(iv) at most one head
Answer:
When two coins are tossed, the sample space of possible outcomes is HH, HT, TH, and TT. The total outcomes count is 4.
(i) Favourable outcomes for exactly one tail are HT and TH (2 outcomes).
\( P(\text{exactly one tail}) = \frac{2}{4} = \frac{1}{2} \)
(ii) Favourable outcomes for at least one head (meaning 1 or 2 heads) are HH, HT, and TH (3 outcomes).
\( P(\text{at least one head}) = \frac{3}{4} \)
(iii) Favourable outcome for no head is TT (1 outcome).
\( P(\text{no head}) = \frac{1}{4} \)
(iv) "At most one head" means we can have 0 or 1 head. The favourable outcomes are TT, HT, and TH (3 outcomes).
\( P(\text{at most one head}) = \frac{3}{4} \)
In simple words: With two coins, four results can happen. Look at each question to see which of the four options work, then write the fraction and simplify it if possible.

Exam Tip: Understand the difference between "at least" (that number or more) and "at most" (that number or less). For example, "at most one head" includes the outcome with zero heads (TT).