CBSE Class 10 Maths Probability Worksheet Set 05

Very Short Answer Questions

Question. A die is thrown once. What is the probability of getting a prime number? 
Answer: When a die is thrown once total number of possible outcomes \( = 6 \)
We have prime numbers on a die \( = \{2, 3, 5\} \)
\( \therefore \) Number of favourable outcomes \( = 3 \)
\( \therefore \) Probability of getting a prime \( = \frac{3}{6} = \frac{1}{2} \).

Question. Find the probability of getting a black queen when a card is drawn at random from a well-shuffled pack of 52 cards. 
Answer: Probability of getting a black queen \( = \frac{2}{52} = \frac{1}{26} \) 1

Question. The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow? 
Answer: Given probability that it will rain tomorrow is 0.85 i.e., \( P = 0.85 \)
\( \therefore \) Probability that it will not rain tomorrow \( = 1 - P = 1 - 0.85 = 0.15 \)

Question. A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen. 
Answer: Total cards \( = 52 \)
Total red cards & queen \( = 28 \)
Probability of getting neither red card nor queen
\( = 1 - \frac{28}{52} = \frac{52 - 28}{52} \)
\( = \frac{24}{52} = \frac{12}{26} = \frac{6}{13} \)
Ans :- \( \frac{12}{26} \) or \( \frac{6}{13} \) 

Question. 20 tickets, on which numbers 1 to 20 are written, are mixed thoroughly and then a ticket is drawn at random out of them. Find the probability that the number on the drawn ticket is a multiple of 3 or 7. 
Answer: \( n(S) = 20 \), multiples of 3 or 7, \( A: \{3, 6, 9, 12, 15, 18, 7, 14\} \), \( n(A) = 8 \)
\( \therefore \) Required probability \( = \frac{8}{20} \) or \( \frac{2}{5} \)

Question. A number is chosen at random from the numbers \( -3, -2, -1, 0, 1, 2, 3 \). What will be the probability that square of this number is less than or equal to 1 ?
Answer: Favourable outcomes are \( -1, 0, 1 = 3 \)
Total outcomes \( = 7 \)
\( \therefore \) Required probability \( = \frac{3}{7} \)

Question. Cards marked with number 3, 4, 5, ....., 50 are placed in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the selected card bears a perfect square number. 
Answer: Possible outcomes are 4, 9, 16, 25, 36, 49, i.e., 6.
\( \therefore P(\text{perfect square number}) = \frac{6}{48} \) or \( \frac{1}{8} \)

Question. Two dice are thrown simultaneously. What is the probability that the sum of the two numbers appearing on the top is 13? 
Answer: When two dice are thrown simultaneously, the maximum sum of the numbers on both the dice is 12, i.e. 6 on each die.
\( \therefore \) Probability of getting sum 13 on both dice \( = 0 \).

Short Answer Questions–I

Question. Cards numbered 7 to 40 were put in a box. Poonam selects a card at random. What is the probability that Poonam selects a card which is a multiple of 7? 
Answer: Total number of possible outcomes \( = 34 \)
Favourable number of outcomes is \( \{7, 14, 21, 28 \text{ and } 35\} = 5 \)
\( P(\text{multiple of } 7) = \frac{5}{34} \) 

Question. A child has a die whose six faces show the letters as shown below:
A B C D E A
The die is thrown once. What is the probability of getting (i) A (ii) D ? 
Answer: When a die is thrown once.
The total number of possible outcomes \( = 6 \)
(i) There are two faces which show a letter A.
\( \therefore \) No. of favourable outcomes \( = 2 \)
\( \therefore \) Probability of getting a letter A \( = \frac{2}{6} = \frac{1}{3} \)
(ii) There is only one face showing D.
\( \therefore \) No. of favourable outcomes \( = 1 \)
\( \therefore \) Probability of getting a letter D \( = \frac{1}{6} \)

Question. Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is a prime number. 
Answer: Product of the number on the dice is prime number, i.e., 2, 3, 5.
The possible ways are \( (1, 2), (2, 1), (1, 3), (3, 1), (5, 1), (1, 5) \).
So, number of favourable ways \( = 6 \)
\( \therefore \) Required probability \( = \frac{6}{36} = \frac{1}{6} \)

Question. Rahim tosses two different coins simultaneously. Find the probability of getting at least one tail.
Answer: The sample space is \( \{HH, HT, TH, TT\} \)
Total number of outcomes \( = 4 \)
Favourable outcomes for getting at least one tail is \( \{HT, TH, TT\} \).
Number of favourable outcomes \( = 3 \)
\( \therefore \) Probability of getting at least one tail \( = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{3}{4} \)

Question. Find the probability that a leap year selected at random will contain 53 Sundays and 53 Mondays.
Answer: We know that a leap year has 366 days.
So in 366 days, there are 52 weeks and 2 days.
Now two days may be
(i) Sunday and Monday
(ii) Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday
Now in total 7 possibilities, Sunday and Monday both come together is only one time
\( \therefore \) Required probability \( = \frac{1}{7} \)

Question. Tree Plantation Drive
A Group Housing Society has 600 members, who have their houses in the campus and decided to hold a tree plantation drive on the occasion of new year. Each household was given the choice of planting a sampling of its choice. The number of different types of samplings planted were :
(i) Neem - 125
(ii) Peepal - 165
(iii) Creepers - 50
(iv) Fruit plants - 150
(v) Flowering plants - 110
On the opening ceremony, one of the plants is selected randomly for a prize. After reading the above passage, answer the following questions.
What is the probability that the selected plant is
(i) a fruit plant or a flowering plant?
(ii) either a Neem plant or a Peepal plant? [Competency Based Question,
Answer: Total number of trees planted \( = 600 \)
\( \therefore \) No. of total possible outcomes \( = 600 \)
(i) The probability that the selected plant is a fruit plant or a flowering plant
\( = \frac{150 + 110}{600} = \frac{260}{600} = \frac{13}{30} \)
(ii) Probability that selected plant is either a Neem plant or a Peepal plant \( = \frac{125 + 165}{600} \)
\( = \frac{290}{600} = \frac{29}{60} \)

Question. An integer is chosen at random between 1 and 100. Find the probability that it is :
(i) divisible by 8. (ii) not divisible by 8.
Answer: Integers, 1 to 100 (between)
\( \implies \) total \( = 98 \) possible outcomes.
i) divisible by \( 8 \implies 12 \) numbers. \( \{8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96\} \).
\( \implies \) Probability \( = \frac{\text{Favourable outcome}}{\text{Total outcome}} = \frac{12}{98} = \frac{6}{49} \).
ii) not divisible by \( 8 \implies 98 - 12 = 86 \) numbers.
\( \implies \) Probability \( = \frac{\text{Favourable outcome}}{\text{Total outcome}} = \frac{86}{98} = \frac{43}{49} \). 

Short Answer Questions–II

Each of the following questions are of 3 marks.

Question. Read the following passage and answer the questions given at the end:
Diwali Fair
A game in a booth at a Diwali Fair involves using a spinner first. Then, if the spinner stops on an even number, the player is allowed to pick a marble from a bag. The spinner and the marbles in the bag are represented in Fig, 14.4. Prizes are given, when a black marble is picked. Shweta plays the game once.
(i) What is the probability that she will be allowed to pick a marble from the bag?
(ii) Suppose she is allowed to pick a marble from the bag, what is the probability of getting a prize, when it is given that the bag contains 20 balls out of which 6 are black?  
Answer: (i) It is clear that if the spinner stops on an even number, then she will be allowed to pick a marble.
\( \therefore \) Probability that she will be allowed to pick a marble from the bag is equal to probability that spinner stops on an even number.
Required probability \( = \frac{5}{6} \)
(ii) Now,
Probability of getting a prize = probability of getting a black marble
\( = \frac{6}{20} = \frac{3}{10} \)

Question. A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag. 
Answer: Let there be \( x \) black balls and 15 white balls.
Total balls \( = n(S) = 15 + x \)
\( P(\text{drawing black ball}) = 3 \times P(\text{drawing white ball}) \).
\( \frac{x}{(15 + x)} = 3 \times \frac{15}{(15 + x)} \)
\( x = 45 \)
\( \therefore \) There are 45 black balls in the bag. 

Question. Two different dice are thrown together. Find the probability that the numbers obtained
(i) have a sum less than 7
(ii) have a product less than 16
(iii) is a doublet of odd numbers
Answer: Total number of outcomes \( = 36 \)
(i) Favourable outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1) i.e., 15.
\( \therefore P(\text{sum less than } 7) = \frac{15}{36} \) or \( \frac{5}{12} \)
(ii) Favourable outcomes are (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), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2) i.e., 25.
\( \therefore P(\text{product less than } 16) = \frac{25}{36} \)
(iii) Favourable outcomes are (1, 1), (3, 3), (5, 5) i.e., 3.
\( \therefore P(\text{doublet of odd number}) = \frac{3}{36} \) or \( \frac{1}{12} \)

Question. A game consists of tossing a coin 3 times and noting the outcome each time. If getting the same result in all the tosses is a success, find the probability of losing the game. 
Answer: Total number of outcomes \( = 8 \)
Favourable number of outcomes (HHH, TTT) \( = 2 \)
Prob. (getting success) \( = \frac{2}{8} \) or \( \frac{1}{4} \)
\( \therefore \) Prob. (losing the game) \( = 1 - \frac{1}{4} = \frac{3}{4} \). 

Question. From a pack of 52 playing cards, Jacks, Queens and Kings of red colour are removed. From the remaining, a card is drawn at random. Find the probability that drawn card is:
(i) a black King (ii) a card of red colour (iii) a card of black colour 
Answer: Total cards \( = 52 \)
Cards removed \( = 6 \)
Card left \( = 52 - 6 = 46 \)
Total black king \( = 2 \).
Probability of drawing black king \( = \frac{2}{46} = \frac{1}{23} \)
Total red card \( = 26 - 6 = 20 \)
Probability of drawing red colour card \( = \frac{20}{46} = \frac{10}{23} \)
Total card of black colour \( = 26 \)
Probability of drawing black colour card \( = \frac{26}{46} = \frac{13}{23} \) 

Question. A die is thrown once. Find the probability of getting a number which (i) is a composite number (ii) lies between 1 and 4.
Answer: Total number of outcomes \( = 6 \).
(i) Prob. (getting a composite number (4, 6)) \( = \frac{2}{6} \) or \( \frac{1}{3} \)
(ii) Prob. (getting a number between 1 and 4 (2, 3)) \( = \frac{2}{6} \) or \( \frac{1}{3} \).

Question. Harpreet tosses two different coins simultaneously (say, one is of Rs. 1 and other of Rs. 2). What is the probability that she gets at least one head?
Answer: When two coins are tossed simultaneously, the possible outcomes are (H, H), (H, T), (T, H), (T, T) which are all equally likely. Here (H, H) means head up on the first coin (say on Rs. 1) and head up on the second coin (Rs. 2). Similarly (H, T) means head up on the first coin and tail up on the second coin and so on.
The outcomes favourable to the event E, 'at least one head' are (H, H), (H, T) and (T, H).
So, the number of outcomes favourable to E is 3.
Therefore, \( P(E) = \frac{3}{4} \)
i.e., the probability that Harpreet gets at least one head is \( \frac{3}{4} \).

Question. A bag contains 5 red, 8 white and 7 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is:
(i) red or white. (ii) not black. (iii) neither white nor black.
Answer: Total number of balls \( = 5 + 8 + 7 = 20 \)
(i) P (red or white ball) \( = \frac{5 + 8}{20} = \frac{13}{20} \)
(ii) P (not black ball) \( = 1 - P(\text{black ball}) = 1 - \frac{7}{20} = \frac{13}{20} \)
(iii) P (neither white nor black ball) \( = P(\text{red ball}) = \frac{5}{20} = \frac{1}{4} \).

Long Answer Questions

Question. Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number 36? Why? 
Answer: Sol. Apoorv throws two dice once.
So, total number of outcomes, \( n(S) = 36 \).
Number of outcomes for getting product 36, \( n(E_1) = 1 [ (6 \times 6) ] \)
\( \therefore \) Probability for Apoorv getting the number 36 \( = \frac{n(E_1)}{n(S)} = \frac{1}{36} \)
Also, Peehu throw one die.
So, total number of outcomes \( n(S) = 6 \)
Number of outcomes for getting square of a number as 36. \( n(E_2) = 1 (\because 6^2 = 36) \)
\( \therefore \) Probability for Peehu getting the number 36 \( = \frac{n(E_2)}{n(S)} = \frac{1}{6} = \frac{6}{36} \)
Hence, Peehu has better chance of getting the number 36.

Question. A number x is selected at random from the numbers 1, 2, 3 and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that product of x and y is less than 16. 
Answer: We have –
Total possible outcome = 1, 2, 3, 4 & 1, 4, 9, 16 = 16
\[ \begin{array}{c|cccc} & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ 4 & 4 & 8 & 12 & 16 \\ 9 & 9 & 18 & 27 & 36 \\ 16 & 16 & 32 & 48 & 64 \end{array} \] Total favourable event, having product less than 16 = 1, 2, 3, 4, 4, 8, 12, 9 = 7 + 1 = 8
Probability = \( \frac{\text{Favourable outcome}}{\text{Total outcome}} \)
\( P(E) = \frac{7 + 1}{16} = \frac{8}{16} = \frac{1}{2} \)
Ans: \( \frac{1}{2} \) 

Question. A bag contains cards numbered from 1 to 49. A card is drawn from the bag at random, after mixing the cards thoroughly. Find the probability that the number on the drawn card is:
(i) an odd number. (ii) a multiple of 5. (iii) a perfect square. (iv) an even prime number. 
Answer: Total number of cards = 49
\( \therefore \) Total number of outcomes = 49
(i) Odd number
Favourable outcomes : 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49 i.e., 25
Probability (E) \( = \frac{\text{No. of favourable outcomes}}{\text{Total number of outcomes}} = \frac{25}{49} \)
(ii) A multiple of 5
Favourable outcomes : 5, 10, 15, 20, 25, 30, 35, 40, 45 i.e., 9
Probability (E) \( = \frac{\text{No. of favourable outcomes}}{\text{Total number of outcomes}} = \frac{9}{49} \)
(iii) A perfect square
Favourable outcomes : 1, 4, 9, 16, 25, 36, 49 i.e., 7
Probability (E) \( = \frac{\text{No. of favourable outcomes}}{\text{Total number of outcomes}} = \frac{7}{49} = \frac{1}{7} \)
(iv) An even prime number
Favourable outcome : 2 i.e., 1
Probability (E) \( = \frac{\text{No. of favourable outcomes}}{\text{Total number of outcomes}} = \frac{1}{49} \)

Question. The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is \( \frac{1}{5} \). The probability of selecting a black marble at random from the same jar is \( \frac{1}{4} \). If the jar contains 11 green marbles, find the total number of marbles in the jar. 
Answer: \( P(\text{blue marble}) = \frac{1}{5}, P(\text{black marble}) = \frac{1}{4} \)
\( \therefore P(\text{green marble}) = 1 - \left( \frac{1}{5} + \frac{1}{4} \right) = \frac{11}{20} \)
Let total number of marbles be \( x \)
Then \( \frac{11}{20} \times x = 11 \)
\( \implies \) \( x = 20 \) 

Case Study-based Questions

Each of the following questions are of 4 marks.

Read the following and answer any four questions from (i) to (v).
On a weekend Rani was playing cards with her family. The deck has 52 cards. If her brother drew one card. 

Question. The probability of getting a king of red colour is
(a) \( \frac{1}{26} \)
(b) \( \frac{1}{13} \)
(c) \( \frac{1}{52} \)
(d) \( \frac{1}{4} \)
Answer: (a) \( \frac{1}{26} \)

Question. The probability of getting a face card is
(a) \( \frac{1}{26} \)
(b) \( \frac{1}{13} \)
(c) \( \frac{2}{13} \)
(d) \( \frac{3}{13} \)
Answer: (d) \( \frac{3}{13} \)

Question. The probability of getting a jack of hearts is
(a) \( \frac{1}{26} \)
(b) \( \frac{1}{52} \)
(c) \( \frac{3}{52} \)
(d) \( \frac{3}{26} \)
Answer: (b) \( \frac{1}{52} \)

Question. The probability of getting a red face card is
(a) \( \frac{3}{26} \)
(b) \( \frac{1}{13} \)
(c) \( \frac{1}{52} \)
(d) \( \frac{1}{4} \)
Answer: (a) \( \frac{3}{26} \)

Question. The probability of getting a spade is
(a) \( \frac{1}{26} \)
(b) \( \frac{1}{13} \)
(c) \( \frac{1}{52} \)
(d) \( \frac{1}{4} \)
Answer: (d) \( \frac{1}{4} \)

Read the following and answer any four questions from (i) to (v).
Rahul and Ravi planned to play Business (board game) in which they were supposed to use two dice. 

Question. Ravi got first chance to roll the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice as 8?
(a) \( \frac{1}{26} \)
(b) \( \frac{5}{36} \)
(c) \( \frac{1}{18} \)
(d) 0
Answer: (b) \( \frac{5}{36} \)

Question. Rahul got next chance. What is the probability that he got the sum of the two numbers appearing on the top face of the dice as 13?
(a) 1
(b) \( \frac{5}{36} \)
(c) \( \frac{1}{18} \)
(d) 0
Answer: (d) 0

Question. Now it was Ravi's turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice less than or equal to 12?
(a) 1
(b) \( \frac{5}{36} \)
(c) \( \frac{1}{18} \)
(d) 0
Answer: (a) 1

Question. Rahul got next chance. What is the probability that he got the sum of the two numbers appearing on the top face of the dice equal to 7?
(a) \( \frac{5}{9} \)
(b) \( \frac{5}{36} \)
(c) \( \frac{1}{6} \)
(d) 0
Answer: (c) \( \frac{1}{6} \)

Question. Now it was Ravi's turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice greater than 8?
(a) 1
(b) \( \frac{5}{36} \)
(c) \( \frac{1}{18} \)
(d) \( \frac{5}{18} \)
Answer: (d) \( \frac{5}{18} \)

Two friends are travelling in a bus. They were feeling bored, so they started playing a game with a pair of dice that one of them had. Each of them started rolling the pair of dice one by one, stating one condition before rolling. If the person gets the numbers according to the condition stated by him, he wins and get a score.

Question. First friend says, "a doublet". What is the probability of his winning?
Answer: Number of doublets are {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} i.e., 6.
Total possible events = 36
\( \therefore P(E) = \frac{6}{36} = \frac{1}{6} \)

Question. Second friend says, "sum less than 9". What is the probability of his winning?
Answer: Possible cases of sum less than 9 are {(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), (4, 1), (4, 2), (4, 3), (4, 4), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2)} i.e., 26.
Total possible events = 36
\( \therefore P(E) = \frac{26}{36} = \frac{13}{18} \)

Question. First one says, "6 will come up either time." What is the probability of his winning?
Answer: Possible cases when 6 will come up either time are {(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} i.e., 11.
Number of favourable outcomes = 11.
\( \therefore P(E) = \frac{11}{36} \)

Question. Second one says, "sum is an even number". What is the probability of his losing?
Answer: Possible cases for which sum is an even number are {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)} i.e., 18.
\( \therefore P(E) = \frac{18}{36} = \frac{1}{2} \)
Probability of his losing is \( 1 - \frac{1}{2} = \frac{1}{2} \)