CBSE Class 10 Maths HOTs Probability Set 04

ASSERTION AND REASON QUESTIONS

In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Choose the correct answer out of the following choices.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question. Assertion (A): Three unbiased coins are tossed together, then the probability of getting exactly 1 head is \( \frac{3}{8} \).
Reason (R): Favourable number of outcomes do not lie in the sample space of total number of out comes.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Answer: (c) A is true but R is false.

Question. Assertion (A): Seven face cards are removed from a deck of cards and the cards are well shuffled. Then the probability of drawing a face card is \( \frac{5}{52} \).
Reason (R): King, Queen and Jack are known as face cards. So, there are 12 face cards in total.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Answer: (d) A is false but R is true.

CASE-BASED QUESTIONS

On a weekend Rani was playing cards with her family. The deck has 52 cards. If her brother drew one card.

Question. Find the probability of getting a king of red colour.

Answer: Total cards = 52. Number of red kings = 2.
Probability = \( \frac{2}{52} = \frac{1}{26} \)

Question. Find the probability of getting a face card.

Answer: Total face cards = 12.
Probability = \( \frac{12}{52} = \frac{3}{13} \)

Question. Find the probability of getting a jack of hearts.

Answer: Number of jack of hearts = 1.
Probability = \( \frac{1}{52} \)

Question. Find the probability of getting a red face card.

Answer: Total red face cards = 6.
Probability = \( \frac{6}{52} = \frac{3}{26} \)

During summer break, Harish wanted to play with his friends but it was too hot outside, so he decided to play some indoor game with his friends. He collects 20 identical cards and writes the numbers 1 to 20 on them (one number on one card). He puts them in a box. He and his friends make a bet for the chances of drawing various cards out of the box. Each was given a chance to tell the probability of picking one card out of the box.

Question. Find the probability that the number on the card drawn is an odd prime number.

Answer: Odd prime numbers from 1 to 20 are 3, 5, 7, 11, 13, 17, 19. Total = 7.
Total cards = 20.
Probability = \( \frac{7}{20} \)

Question. Find the probability that the number on the card drawn is a composite number.

Answer: Composite numbers from 1 to 20 are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20. Total = 11.
Probability = \( \frac{11}{20} \)

Question. Find the probability that the number on the card drawn is a multiple of 3, 6 and 9.

Answer: Multiples of 3 are 3, 6, 9, 12, 15, 18.
Multiples of 6 are 6, 12, 18.
Multiples of 9 are 9, 18.
The only number that is a multiple of all three (LCM is 18) in the range 1-20 is 18. Total = 1.
Probability = \( \frac{1}{20} \)

Question. If all cards having odd numbers written on them are removed from the box and then one card is drawn from the remaining cards, then find probability of getting a card having a prime number.

Answer: Total cards initially = 20. Odd numbers (1, 3, 5, 7, 9, 11, 13, 15, 17, 19) are removed. Total removed = 10.
Remaining cards = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}. Total remaining = 10.
Prime numbers in remaining cards = {2}. Total = 1.
Probability = \( \frac{1}{10} \)

HOOK-A-DUCK is a traditional stall game. A number of identical rubber ducks are floated in a water trough. The ducks have metal rings fastened to their heads. The player captures the ducks using a pole with a hook at one end. Although the ducks appear identical, the letters S, M or L appears on the bottom of some ducks, signifying that the player who captures them receives a Small, Medium or Large prize respectively. A duck with no letter signifies no prize. In order to play the game, a player needs to buy a token and is given three attempts per token. In each attempt, he needs to capture a duck. The captured ducks are examined to reveal whether they have any letter and the player gets the prizes accordingly. The ducks are returned to the trough only after the player finishes all the three attempts. If a player buys more than one token, the ducks are returned to the trough after every 3 attempts. Sam is playing one such game where there are 120 ducks in the trough. The ratio of the number of ducks without any letter to the number of ducks with letters S, M and L is 10 : 6 : 3 : 1 respectively.

Question. Sam is playing his first attempt of the game. What is the probability that he wins a medium prize?

Answer: Total ratio = 10 + 6 + 3 + 1 = 20.
Total ducks = 120.
Number of ducks for medium prize (M) = \( \frac{3}{20} \times 120 = 18 \).
Probability = \( \frac{18}{120} = \frac{3}{20} \)

Question. A player captures empty ducks in his first two attempts. What is the probability that he will win a small prize in his third attempt?

Answer: Total ducks = 120. Two empty ducks are already removed and not returned yet.
Remaining ducks = 118.
Number of ducks for small prize (S) = \( \frac{6}{20} \times 120 = 36 \).
Probability = \( \frac{36}{118} = \frac{18}{59} \)

Question. Mera is playing the game and she got a duck marked S in her first attempt. Which duck will be most likely captured by her in the second attempt?

Answer: After picking an 'S', the distribution is:
No letter: 60 ducks
S: 35 ducks remaining
M: 18 ducks
L: 6 ducks
The most likely duck to be captured is the one with 'No letter' as it has the highest count (60).

Question. A certain number of new ducks, 18 of which marked L when added to the trough, doubles the probability of winning a large prize in the first attempt. Find the number of ducks added.

Answer: Initial number of L ducks = 6. Initial total = 120. Initial \( P(L) = \frac{6}{120} = \frac{1}{20} \).
Let total ducks added be \( n \). New L ducks = \( 6 + 18 = 24 \).
New total ducks = \( 120 + n \).
New probability \( P'(L) = \frac{24}{120 + n} \).
According to the condition: \( P'(L) = 2 \times P(L) \)
\( \implies \) \( \frac{24}{120 + n} = 2 \times \frac{6}{120} \)
\( \implies \) \( \frac{24}{120 + n} = \frac{12}{120} \)
\( \implies \) \( \frac{2}{120 + n} = \frac{1}{120} \)
\( \implies \) \( 120 + n = 240 \)
\( \implies \) \( n = 120 \).
Total ducks added = 120.

Question. Cards bearing numbers 2, 3, 4, ..., 11 are kept in a bag. A card is drawn at random from the bag. The probability of getting a card with a prime number is
(a) \( \frac{1}{2} \)
(b) \( \frac{2}{5} \)
(c) \( \frac{3}{10} \)
(d) \( \frac{5}{9} \)

Answer: (a) \( \frac{1}{2} \)

Question. There are 25 tickets bearing numbers from 1 to 25. One ticket is drawn at random. The probability that the number on it is a multiple of 5 or 6 is
(a) \( \frac{7}{25} \)
(b) \( \frac{9}{25} \)
(c) \( \frac{11}{12} \)
(d) \( \frac{13}{25} \)

Answer: (b) \( \frac{9}{25} \)

Question. The probability of getting a number between 1 and 100 which is divisible by 1 and itself only is
(a) \( \frac{29}{98} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{25}{98} \)
(d) \( \frac{23}{98} \)

Answer: (c) \( \frac{25}{98} \)

Question. A single letter is selected at random from the word ‘PROBABILITY’. The probability that it is a vowel is
(a) \( \frac{3}{11} \)
(b) \( \frac{4}{11} \)
(c) \( \frac{2}{11} \)
(d) \( \frac{5}{11} \)

Answer: (b) \( \frac{4}{11} \)

Question. Among 52 cards, there are 12 face cards. Probability that a card drawn at random is not a face card is
(a) \( \frac{3}{13} \)
(b) \( \frac{9}{13} \)
(c) \( \frac{10}{13} \)
(d) \( \frac{3}{4} \)

Answer: (c) \( \frac{10}{13} \)

Question. A box contains cards numbered 6 to 50. A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect square, is
(a) \( \frac{1}{9} \)
(b) \( \frac{2}{9} \)
(c) \( \frac{4}{9} \)
(d) \( \frac{5}{9} \)

Answer: (a) \( \frac{1}{9} \)

Question. The king, queen and jack of clubs are removed from a deck of 52 playing cards and the cards are well shuffled. One card is selected from the remaining cards. Find the probability of getting a heart.

Answer: Total cards remaining = 52 - 3 = 49.
Number of heart cards = 13.
Probability = \( \frac{13}{49} \)

Question. Two dice are thrown simultaneously. Find the probability that the sum of the two numbers appearing on the top is less than or equal to 10.

Answer: Total outcomes = 36. Outcomes with sum > 10 are (5,6), (6,5), (6,6). Total = 3.
Favourable outcomes = 36 - 3 = 33.
Probability = \( \frac{33}{36} = \frac{11}{12} \)

Question. The king, queen and jack of diamonds are removed from a pack of 52 cards and then the pack is well shuffled. A card is drawn from the remaining cards. find the probability of getting a card of (i) diamonds, (ii) a jack.

Answer: Total cards remaining = 49.
(i) Number of diamonds remaining = 13 - 3 = 10. Probability = \( \frac{10}{49} \)
(ii) Number of jacks remaining = 4 - 1 = 3. Probability = \( \frac{3}{49} \)

Question. A bag contains cards which are numbered from 2 to 90. A card is drawn at random from the bag. Find the probability that it bears (a) a two digit number, (b) a number which is a perfect square.

Answer: Total numbers from 2 to 90 = 90 - 2 + 1 = 89.
(a) Two digit numbers are from 10 to 90. Count = 90 - 10 + 1 = 81. Probability = \( \frac{81}{89} \)
(b) Perfect squares are 4, 9, 16, 25, 36, 49, 64, 81. Count = 8. Probability = \( \frac{8}{89} \)

Question. Cards marked with numbers 1, 3, 5, ..., 101 are placed in a bag and mixed thoroughly. A card is then drawn at random from the bag. Find the probability that the number on the drawn card is (i) less than 19. (ii) a prime number less than 20.

Answer: Total numbers = \( \frac{101 - 1}{2} + 1 = 51 \).
(i) Numbers less than 19 are 1, 3, 5, 7, 9, 11, 13, 15, 17. Count = 9. Probability = \( \frac{9}{51} = \frac{3}{17} \)
(ii) Prime numbers less than 20 are 3, 5, 7, 11, 13, 17, 19. Count = 7. Probability = \( \frac{7}{51} \)

Question. Three coins are tossed. Find the probability of
(i) getting exactly one head
(ii) getting at least one head and one tail.

Answer: Total outcomes = 8.
(i) Exactly one head: {HTT, THT, TTH}. Probability = \( \frac{3}{8} \)
(ii) At least one head and one tail: {HHT, HTH, THH, HTT, THT, TTH}. Probability = \( \frac{6}{8} = \frac{3}{4} \)

Question. A box contains cards numbered 3, 5, 7, 9, ..., 35, 37. A card is drawn at random from the box. Find the probability that the number on the drawn card is a prime number.

Answer: Total cards = \( \frac{37 - 3}{2} + 1 = 18 \).
Prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37. Count = 11.
Probability = \( \frac{11}{18} \)

Question. If odds in favour of an event be 2 : 3, find the probability of non-occurrence of this event.

Answer: Odds in favour = 2 : 3.
Probability of occurrence = \( \frac{2}{2+3} = \frac{2}{5} \).
Probability of non-occurrence = \( 1 - \frac{2}{5} = \frac{3}{5} \)

Question. A box has cards numbered 114 to 199. cards are mixed thoroughly and a card is drawn from the bag at random. Find the probability that the number on the card, drawn from the box is (a) an odd number (b) a perfect square number

Answer: Total cards = 199 - 114 + 1 = 86.
(a) Half of the numbers are odd, half are even. Number of odd cards = 43. Probability = \( \frac{43}{86} = \frac{1}{2} \)
(b) Perfect squares between 114 and 199 are \( 11^2=121, 12^2=144, 13^2=169, 14^2=196 \). Count = 4. Probability = \( \frac{4}{86} = \frac{2}{43} \)

Question. If odds against an event be 3 : 4, find the probability of occurrence of this event.

Answer: Odds against = 3 : 4.
Probability of non-occurrence = \( \frac{3}{3+4} = \frac{3}{7} \).
Probability of occurrence = \( 1 - \frac{3}{7} = \frac{4}{7} \)

Question. Find the probability that the month of June has exactly 5 Tuesdays.

Answer: June has 30 days = 4 weeks + 2 days. The 5th Tuesday occurs if either of the 2 extra days is Tuesday.
Possible pairs of extra days: (M, T), (T, W), (W, Th), (Th, F), (F, S), (S, Su), (Su, M). Total = 7.
Favourable cases are (M, T) and (T, W). Count = 2.
Probability = \( \frac{2}{7} \)

Question. A bag contains 30 balls out of which x are black. If 10 more black balls are put in the box. The probability of drawing a black ball is double of what it was before. Find the value of x.

Answer: Initial \( P(\text{black}) = \frac{x}{30} \).
New total = 40. New black balls = x + 10.
New \( P'(\text{black}) = \frac{x + 10}{40} \).
According to the condition: \( P'(\text{black}) = 2 \times P(\text{black}) \)
\( \implies \) \( \frac{x + 10}{40} = 2 \times \frac{x}{30} = \frac{x}{15} \)
\( \implies \) \( 15(x + 10) = 40x \)
\( \implies \) \( 15x + 150 = 40x \)
\( \implies \) \( 25x = 150 \)
\( \implies \) \( x = 6 \)

Question. A bag contains 15 red balls and some green balls. If probability of drawing a green ball is three times that of a red ball, determine the number of green balls in the bag.

Answer: Let green balls be \( g \). Total = 15 + \( g \).
\( P(\text{green}) = 3 \times P(\text{red}) \)
\( \implies \) \( \frac{g}{15+g} = 3 \times \frac{15}{15+g} \)
\( \implies \) \( g = 45 \)

Question. A box contains 40 balls out of which x are white. If one ball is drawn at random from the bag what is the probability it is white. If 20 more white balls are put in the box, the probability of drawing white ball is now same what it was before. Find x.

Answer: Initial probability = \( \frac{x}{40} \).
New probability = \( \frac{x+20}{60} \).
Condition: \( \frac{x}{40} = \frac{x+20}{60} \)
\( \implies \) \( 60x = 40(x+20) \)
\( \implies \) \( 6x = 4x + 80 \)
\( \implies \) \( 2x = 80 \)
\( \implies \) \( x = 40 \)

Question. A box contains 50 marbles, some are red and others are green. If a marble is drawn from the box, the probability it is red is \( \frac{2}{5} \). Find the number of green marbles is the box.

Answer: Total marbles = 50.
Number of red marbles = \( \frac{2}{5} \times 50 = 20 \).
Number of green marbles = 50 - 20 = 30.

Direction (Q. 22 and 23): In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Choose the correct answer out of the following choices.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question. Assertion (A): Cards numbered 5 to 102 are placed in a box. If a card is selected at random from the box, then the probability that the card selected has a number which is a perfect square is \( \frac{4}{49} \).
Reason (R): Probability of an event E is a number such that 0 ≤ P(E) ≤ 1.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Answer: (b) Both A and R are true but R is not the correct explanation of A.

Question. Assertion (A): In a game, the entry fee is Rs. 10. The game consists of tossing of 3 coins. If one or two heads show Amita wins the game and gets entry fee. The probability, that she gets the entry fee is \( \frac{3}{4} \).
Reason (R): When three coins are tossed together, all the outcomes are {HHH, HHT, HTH, THH, HTT, THT, TTH and TTT}
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Answer: (a) Both A and R are true and R is the correct explanation of A.

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?

Answer: Outcomes with sum 8 are (2,6), (3,5), (4,4), (5,3), (6,2). Count = 5.
Probability = \( \frac{5}{36} \)

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?

Answer: Maximum sum possible with two dice is 12 (6,6). A sum of 13 is impossible.
Probability = 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?

Answer: All possible outcomes (36 in total) have a sum between 2 and 12 inclusive.
Probability = \( \frac{36}{36} = 1 \)