CBSE Class 10 Mathematics Probability MCQs Set 11

Mathematics Objective Questions and Answers: Chapter 14 Probability

Review structured MCQ sets for Class 10 Mathematics Chapter 14 Probability. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Download Chapter 14 Probability MCQs with Answers

Navigate directly to the 50 objective questions for Chapter 14 Probability using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: In a family of three children, the probability of having at least one boy is
(a) \( \frac{7}{8} \)
(b) \( \frac{1}{8} \)
(c) \( \frac{5}{8} \)
(d) \( \frac{3}{4} \)
Show Answer & Explanation

Answer: (a) \( \frac{7}{8} \)

Explanation:
1. Each child is a boy or a girl, so there are \( 2 \times 2 \times 2 = 8 \) equally likely outcomes: BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG.
2. Only GGG has no boy, so P(at least one boy) = \( 1 - \frac{1}{8} = \frac{7}{8} \).

Teacher's Note:
1. Count the equally likely outcomes in order; GGB and GBG are different outcomes.
2. For 'at least one', use 1 − P(none).

Question 2: The probability that a number selected at random from the numbers 1, 2, 3, ..., 15 is a multiple of 4 is
(a) \( \frac{4}{15} \)
(b) \( \frac{2}{15} \)
(c) \( \frac{1}{5} \)
(d) \( \frac{1}{3} \)
Show Answer & Explanation

Answer: (c) \( \frac{1}{5} \)

Explanation:
1. Total outcomes = 15. Multiples of 4 are 4, 8, 12, so 3 are favourable.
2. P = \( \frac{3}{15} = \frac{1}{5} \).

Teacher's Note:
1. List the favourable numbers.

Question 3: A bag contains cards numbered 1 to 25. A card is drawn at random. The probability that the number on the card is divisible by both 2 and 3 is
(a) \( \frac{1}{5} \)
(b) \( \frac{3}{25} \)
(c) \( \frac{4}{25} \)
(d) \( \frac{2}{25} \)
Show Answer & Explanation

Answer: (c) \( \frac{4}{25} \)

Explanation:
1. Divisible by both 2 and 3 means divisible by 6: 6, 12, 18, 24, which are 4 numbers.
2. P = \( \frac{4}{25} \).

Teacher's Note:
1. Divisible by 2 and 3 means divisible by their LCM 6.

Question 4: The probability of getting an even number when a die is thrown once is
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{1}{6} \)
(d) \( \frac{5}{6} \)
Show Answer & Explanation

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

Explanation:
1. Outcomes: 1, 2, 3, 4, 5, 6. Even numbers: 2, 4, 6.
2. P = \( \frac{3}{6} = \frac{1}{2} \).

Teacher's Note:
1. Count even faces.

Question 5: A box contains 90 discs numbered from 1 to 90. If one disc is drawn at random, the probability that it bears a prime number less than 23 is
(a) \( \frac{7}{90} \)
(b) \( \frac{10}{90} \)
(c) \( \frac{4}{45} \)
(d) \( \frac{9}{89} \)
Show Answer & Explanation

Answer: (c) \( \frac{4}{45} \)

Explanation:
1. Primes less than 23 are 2, 3, 5, 7, 11, 13, 17, 19, which are 8.
2. P = \( \frac{8}{90} = \frac{4}{45} \).

Teacher's Note:
1. 1 is not prime.

Question 6: If P(A) denotes the probability of an event A, then
(a) P(A) < 0
(b) P(A) > 1
(c) 0 ≤ P(A) ≤ 1
(d) −1 ≤ P(A) ≤ 1
Show Answer & Explanation

Answer: (c) 0 ≤ P(A) ≤ 1

Explanation:
1. The probability of an impossible event is 0 and of a sure event is 1, and every other event lies between them.
2. So \( 0 \le P(A) \le 1 \).

Teacher's Note:
1. A probability can never be negative or exceed 1.

Question 7: 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 7 is
(a) \( \frac{1}{7} \)
(b) \( \frac{1}{8} \)
(c) \( \frac{1}{5} \)
(d) \( \frac{7}{40} \)
Show Answer & Explanation

Answer: (b) \( \frac{1}{8} \)

Explanation:
1. Multiples of 7 up to 40: 7, 14, 21, 28, 35, which are 5.
2. P = \( \frac{5}{40} = \frac{1}{8} \).

Teacher's Note:
1. Divide 40 by 7 and take the whole part.

Question 8: Two dice are thrown together. The probability of getting the same number on both dice is
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{1}{6} \)
(d) \( \frac{1}{12} \)
Show Answer & Explanation

Answer: (c) \( \frac{1}{6} \)

Explanation:
1. Total outcomes = 36. Favourable outcomes: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), which are 6.
2. P = \( \frac{6}{36} = \frac{1}{6} \).

Teacher's Note:
1. Doublets are the outcomes with equal numbers.

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

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

Explanation:
1. Total cards = 10. Prime numbers: 2, 3, 5, 7, 11, which are 5.
2. P = \( \frac{5}{10} = \frac{1}{2} \).

Teacher's Note:
1. Count the primes from 2 to 11.

Question 10: The probability of throwing a number greater than 2 with a single die is
(a) \( \frac{2}{3} \)
(b) \( \frac{5}{6} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{2}{5} \)
Show Answer & Explanation

Answer: (a) \( \frac{2}{3} \)

Explanation:
1. Favourable numbers: 3, 4, 5, 6, which are 4 out of 6.
2. P = \( \frac{4}{6} = \frac{2}{3} \).

Teacher's Note:
1. Greater than 2 does not include 2.

Question 11: A card is drawn from a well-shuffled deck of 52 playing cards. The probability that the card will not be an ace is
(a) \( \frac{1}{13} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{12}{13} \)
(d) \( \frac{3}{4} \)
Show Answer & Explanation

Answer: (c) \( \frac{12}{13} \)

Explanation:
1. There are 4 aces, so 48 cards are not aces.
2. P = \( \frac{48}{52} = \frac{12}{13} \).

Teacher's Note:
1. Use P(not E) = 1 − P(E) = 1 − 1/13.

Question 12: Which of the following cannot be the probability of an event?
(a) 1.5
(b) \( \frac{3}{5} \)
(c) 1/4
(d) 0.3
Show Answer & Explanation

Answer: (a) 1.5

Explanation:
1. A probability always lies between 0 and 1, so 1.5 is impossible.
2. The others, 3/5, 1/4 and 0.3, all lie between 0 and 1.

Teacher's Note:
1. Check each value against 0 ≤ P ≤ 1.

Question 13: A die is thrown once. The probability of getting a number less than 3 is
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{1}{6} \)
(d) \( \frac{1}{4} \)
Show Answer & Explanation

Answer: (b) \( \frac{1}{3} \)

Explanation:
1. Favourable outcomes: 1 and 2, which are 2 out of 6.
2. P = \( \frac{2}{6} = \frac{1}{3} \).

Teacher's Note:
1. Less than 3 means 1 and 2.

Question 14: Cards bearing the numbers 3 to 20 are placed in a bag and mixed thoroughly. A card is taken out at random. The probability that the number on the card is an even number is
(a) \( \frac{9}{17} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{5}{9} \)
(d) \( \frac{7}{18} \)
Show Answer & Explanation

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

Explanation:
1. Numbers 3 to 20 give 18 cards. Even numbers: 4, 6, 8, ..., 20, which are 9.
2. P = \( \frac{9}{18} = \frac{1}{2} \).

Teacher's Note:
1. The count of numbers from 3 to 20 is 20 − 3 + 1 = 18.

Question 15: A bag contains 4 red and 6 black balls. A ball is taken out of the bag at random. The probability of getting a black ball is
(a) \( \frac{2}{5} \)
(b) \( \frac{3}{5} \)
(c) \( \frac{1}{10} \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( \frac{3}{5} \)

Explanation:
1. Total balls = 10 and black balls = 6.
2. P = \( \frac{6}{10} = \frac{3}{5} \).

Teacher's Note:
1. Favourable over total.

Question 16: A card is drawn from a well-shuffled deck of 52 cards. The probability of getting a black king is
(a) \( \frac{1}{13} \)
(b) \( \frac{1}{26} \)
(c) \( \frac{1}{28} \)
(d) \( \frac{2}{13} \)
Show Answer & Explanation

Answer: (b) \( \frac{1}{26} \)

Explanation:
1. There are 2 black kings (king of spades and king of clubs).
2. P = \( \frac{2}{52} = \frac{1}{26} \).

Teacher's Note:
1. Black cards are spades and clubs.

Question 17: Two friends were born in the year 2000. The probability that they have the same birthday is
(a) \( \frac{1}{365} \)
(b) \( \frac{1}{366} \)
(c) \( \frac{2}{365} \)
(d) \( \frac{1}{183} \)
Show Answer & Explanation

Answer: (b) \( \frac{1}{366} \)

Explanation:
1. The year 2000 is a leap year, so a birthday can be any of 366 days.
2. The second friend's birthday must match the first friend's: 1 favourable day out of 366.
3. P = \( \frac{1}{366} \).

Teacher's Note:
1. Count 366 days for a leap year.

Question 18: The probability that two friends have different birthdays (ignoring leap years) is
(a) \( \frac{1}{365} \)
(b) \( \frac{2}{365} \)
(c) \( \frac{364}{365} \)
(d) \( \frac{363}{365} \)
Show Answer & Explanation

Answer: (c) \( \frac{364}{365} \)

Explanation:
1. Take 365 days in a year. P(same birthday) = \( \frac{1}{365} \).
2. P(different birthdays) = \( 1 - \frac{1}{365} = \frac{364}{365} \).

Teacher's Note:
1. Use P(not E) = 1 − P(E).

Question 19: A die is thrown once. The probability of getting a prime number is
(a) \( \frac{2}{3} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{1}{6} \)
Show Answer & Explanation

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

Explanation:
1. Prime numbers on a die: 2, 3, 5, which are 3 of 6.
2. P = \( \frac{3}{6} = \frac{1}{2} \).

Teacher's Note:
1. 1 is not a prime number.

Question 20: The probability of an impossible event is
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{1} \)
(c) \( \frac{0}{1} \)
(d) \( \frac{2}{1} \)
Show Answer & Explanation

Answer: (c) \( \frac{0}{1} \)

Explanation:
1. An impossible event has no favourable outcome, so its probability is 0.

Teacher's Note:
1. A sure event has probability 1.

Question 21: The probability of getting a number four or more when a die is thrown is
(a) \( \frac{2}{3} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{1}{4} \)
Show Answer & Explanation

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

Explanation:
1. Favourable numbers: 4, 5, 6, which are 3 of 6.
2. P = \( \frac{3}{6} = \frac{1}{2} \).

Teacher's Note:
1. Four or more includes 4.

Question 22: The probability of an event that is certain to happen is
(a) −1
(b) 1
(c) 0
(d) none of these
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. A sure event always occurs, so its probability is 1.

Teacher's Note:
1. Sure event 1, impossible event 0.

Question 23: A number is selected at random from the first 50 natural numbers. The probability of getting a perfect square is
(a) \( \frac{7}{50} \)
(b) \( \frac{1}{50} \)
(c) \( \frac{9}{50} \)
(d) \( \frac{11}{50} \)
Show Answer & Explanation

Answer: (a) \( \frac{7}{50} \)

Explanation:
1. Perfect squares up to 50: 1, 4, 9, 16, 25, 36, 49, which are 7.
2. P = \( \frac{7}{50} \).

Teacher's Note:
1. Squares of 1 to 7 are at most 49.

Question 24: The sum of the probabilities of all the elementary events of an experiment is
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{0}{1} \)
(d) \( \frac{1}{1} \)
Show Answer & Explanation

Answer: (d) \( \frac{1}{1} \)

Explanation:
1. The probabilities of all elementary events of an experiment add up to 1.

Teacher's Note:
1. Think of it as 100% of the sample space.

Question 25: In a game at a fair the entry fee is Rs. 10. The game consists of tossing a coin 3 times. If one or two heads show, Pooja gets her entry fee back. If she throws 3 heads she gets double the entry fee. Otherwise she loses. The probability that she loses the entry fee is
(a) \( \frac{1}{8} \)
(b) \( \frac{3}{8} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{3}{4} \)
Show Answer & Explanation

Answer: (a) \( \frac{1}{8} \)

Explanation:
1. The 8 equally likely outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
2. She loses only with no head, that is TTT: P = \( \frac{1}{8} \).

Teacher's Note:
1. Losing means zero heads.

Question 26: In a game at a fair the entry fee is Rs. 10. The game consists of tossing a coin 3 times. If one or two heads show, Pooja gets her entry fee back. If she throws 3 heads she gets double the entry fee. Otherwise she loses. The probability that she gets double the entry fee is
(a) \( \frac{1}{8} \)
(b) \( \frac{3}{8} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{3}{4} \)
Show Answer & Explanation

Answer: (a) \( \frac{1}{8} \)

Explanation:
1. She needs 3 heads: only HHH out of 8 outcomes.
2. P = \( \frac{1}{8} \).

Teacher's Note:
1. Double the fee means 3 heads.

Question 27: In a game at a fair the entry fee is Rs. 10. The game consists of tossing a coin 3 times. If one or two heads show, Pooja gets her entry fee back. If she throws 3 heads she gets double the entry fee. Otherwise she loses. The probability that she just gets her entry fee back is
(a) \( \frac{1}{8} \)
(b) \( \frac{3}{8} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{3}{4} \)
Show Answer & Explanation

Answer: (d) \( \frac{3}{4} \)

Explanation:
1. One or two heads: HHT, HTH, HTT, THH, THT, TTH, which are 6 outcomes.
2. P = \( \frac{6}{8} = \frac{3}{4} \).

Teacher's Note:
1. Add up outcomes with exactly one or two heads.

Question 28: A ludo board has a square of side 30 cm with a circular portion of diameter 14 cm in the middle. A die is thrown on the board and always lands on the board, and every point of the board is equally likely. The probability that the die lands inside the circular portion is
(a) \( \frac{77}{450} \)
(b) \( \frac{7}{45} \)
(c) \( \frac{11}{50} \)
(d) \( \frac{22}{225} \)
Show Answer & Explanation

Answer: (a) \( \frac{77}{450} \)

Explanation:
1. Area of circle = \( \pi r^2 = \frac{22}{7} \times 7 \times 7 = 154 \) cm².
2. Area of board = 30 × 30 = 900 cm².
3. P = \( \frac{154}{900} = \frac{77}{450} \).

Teacher's Note:
1. Geometric probability is area of the favourable region over total area.

Question 29: What is the probability of getting a prime number on a single throw of a die?
(a) \( \frac{1}{3} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{2}{3} \)
(d) \( \frac{1}{6} \)
Show Answer & Explanation

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

Explanation:
1. Primes on a die: 2, 3, 5, which are 3 of 6.
2. P = \( \frac{3}{6} = \frac{1}{2} \).

Teacher's Note:
1. 1 is not prime.

Question 30: In a class there are 20 girls and 18 boys. The teacher writes the name of each student on a card, mixes the cards in a basket and a child picks one card. The probability that the name on the card is the name of a girl is
(a) \( \frac{10}{19} \)
(b) \( \frac{9}{19} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{20}{18} \)
Show Answer & Explanation

Answer: (a) \( \frac{10}{19} \)

Explanation:
1. Total students = 38 and girls = 20.
2. P = \( \frac{20}{38} = \frac{10}{19} \).

Teacher's Note:
1. Total = 20 + 18.

Question 31: In a class there are 20 girls and 18 boys. The teacher writes the name of each student on a card, mixes the cards in a basket and a child picks one card. The probability that the name on the card is the name of a boy is
(a) \( \frac{10}{19} \)
(b) \( \frac{9}{19} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{18}{20} \)
Show Answer & Explanation

Answer: (b) \( \frac{9}{19} \)

Explanation:
1. Boys = 18 out of 38.
2. P = \( \frac{18}{38} = \frac{9}{19} \).

Teacher's Note:
1. Check: 10/19 + 9/19 = 1.

Question 32: A lot consists of 50 mobile phones, of which 44 are good, 3 have only minor defects and 3 have major defects. Vanshika will buy a phone only if it is good, but the trader will buy a phone only if it has no major defect. One phone is selected at random from the lot. The probability that the phone is acceptable to Vanshika is
(a) \( \frac{22}{25} \)
(b) \( \frac{47}{50} \)
(c) \( \frac{3}{50} \)
(d) \( \frac{1}{2} \)
Show Answer & Explanation

Answer: (a) \( \frac{22}{25} \)

Explanation:
1. Acceptable to Vanshika: 44 good phones.
2. P = \( \frac{44}{50} = \frac{22}{25} \).

Teacher's Note:
1. Only good phones are acceptable to her.

Question 33: A lot consists of 50 mobile phones, of which 44 are good, 3 have only minor defects and 3 have major defects. Vanshika will buy a phone only if it is good, but the trader will buy a phone only if it has no major defect. One phone is selected at random from the lot. The probability that the phone is acceptable to the trader is
(a) \( \frac{22}{25} \)
(b) \( \frac{47}{50} \)
(c) \( \frac{3}{50} \)
(d) \( \frac{1}{2} \)
Show Answer & Explanation

Answer: (b) \( \frac{47}{50} \)

Explanation:
1. The trader accepts good phones and those with only minor defects: 44 + 3 = 47.
2. P = \( \frac{47}{50} \).

Teacher's Note:
1. No major defect means good or minor defect.

Question 34: A lot consists of 50 mobile phones, of which 44 are good, 3 have only minor defects and 3 have major defects. Vanshika will buy a phone only if it is good, but the trader will buy a phone only if it has no major defect. One phone is selected at random from the lot. The probability that the phone is acceptable to neither the trader nor Vanshika is
(a) \( \frac{22}{25} \)
(b) \( \frac{47}{50} \)
(c) \( \frac{3}{50} \)
(d) \( \frac{1}{2} \)
Show Answer & Explanation

Answer: (c) \( \frac{3}{50} \)

Explanation:
1. Phones with major defects are acceptable to neither: 3.
2. P = \( \frac{3}{50} \).

Teacher's Note:
1. Phones rejected by both are those with major defects.

Question 35: In a target game there are three concentric circles of radii 3 cm, 5 cm and 9 cm. A person wins if the dart hits the shaded region between the circles of radii 3 cm and 5 cm, gets a second chance if the dart hits the innermost circle, and loses if the dart hits the outermost ring. The dart always hits the target and every point is equally likely. The probability that the person wins the game is
(a) \( \frac{16}{81} \)
(b) \( \frac{1}{9} \)
(c) \( \frac{56}{81} \)
(d) \( \frac{25}{81} \)
Show Answer & Explanation

Answer: (a) \( \frac{16}{81} \)

Explanation:
1. Area of the shaded region = \( \pi(5^2 - 3^2) = 16\pi \). Area of the whole target = \( \pi \times 9^2 = 81\pi \).
2. P(win) = \( \frac{16\pi}{81\pi} = \frac{16}{81} \).

Teacher's Note:
1. Use areas of circles.

Question 36: In a target game there are three concentric circles of radii 3 cm, 5 cm and 9 cm. A person wins if the dart hits the shaded region between the circles of radii 3 cm and 5 cm, gets a second chance if the dart hits the innermost circle, and loses if the dart hits the outermost ring. The dart always hits the target and every point is equally likely. The probability that the person gets the second chance is
(a) \( \frac{16}{81} \)
(b) \( \frac{1}{9} \)
(c) \( \frac{56}{81} \)
(d) \( \frac{25}{81} \)
Show Answer & Explanation

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

Explanation:
1. Area of the innermost circle = \( 9\pi \).
2. P = \( \frac{9\pi}{81\pi} = \frac{1}{9} \).

Teacher's Note:
1. Innermost circle has radius 3 cm.

Question 37: In a target game there are three concentric circles of radii 3 cm, 5 cm and 9 cm. A person wins if the dart hits the shaded region between the circles of radii 3 cm and 5 cm, gets a second chance if the dart hits the innermost circle, and loses if the dart hits the outermost ring. The dart always hits the target and every point is equally likely. The probability that the person loses the game is
(a) \( \frac{16}{81} \)
(b) \( \frac{1}{9} \)
(c) \( \frac{56}{81} \)
(d) \( \frac{25}{81} \)
Show Answer & Explanation

Answer: (c) \( \frac{56}{81} \)

Explanation:
1. Area of the outer ring = \( \pi(9^2 - 5^2) = 56\pi \).
2. P = \( \frac{56\pi}{81\pi} = \frac{56}{81} \).

Teacher's Note:
1. The three probabilities 16/81, 9/81 and 56/81 add up to 1.

Assertion–Reason Questions

Question 38:
Assertion (A): In a cricket match a batsman hits a boundary 9 times out of 45 balls he plays. The probability that in a given throw he does not hit a boundary is \( \frac{4}{5} \).
Reason (R): P(E) + P(not 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.
Show Answer & Explanation

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

Explanation:
1. Reason: the probability of an event and of its complement add up to 1, so R is true.
2. P(boundary) = \( \frac{9}{45} = \frac{1}{5} \), so P(no boundary) = \( 1 - \frac{1}{5} = \frac{4}{5} \). A is true and is obtained using R: option (a).

Teacher's Note:
1. Use P(not E) = 1 − P(E) for 'does not happen' probabilities.

Question 39:
Assertion (A): The probability of a sure event is 1.
Reason (R): Let E be an event. Then \( 0 \le P(E) \le 1 \).
(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.
Show Answer & Explanation

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

Explanation:
1. Assertion: the entire sample space is a sure event, so its probability is 1. True.
2. Reason: for any event E, \( 0 \le P(E) \le 1 \). True.
3. The Reason only gives the range of probabilities; it does not explain why a sure event has probability 1. Option (b).

Teacher's Note:
1. Both statements can be true without one explaining the other.

Question 40:
Assertion (A): In a group of 3 students the probability of 2 students not having the same birthday is 0.992. Then the probability that the 2 students have the same birthday is 0.128.
Reason (R): If n(A) = 1 and n(S) = 13, then \( P(A) = \frac{1}{13} \).
(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.
Show Answer & Explanation

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

Explanation:
1. P(same birthday) = 1 − 0.992 = 0.008, not 0.128. The Assertion is false.
2. P(A) = \( \frac{n(A)}{n(S)} = \frac{1}{13} \). The Reason is true.
3. Assertion false, Reason true: option (d).

Teacher's Note:
1. Use P(E) = 1 − P(not E).

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FAQs

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