CBSE Class 10 Mathematics Probability MCQs Set 20

Multiple Choice Questions (MCQs) for Class 10 Mathematics: Chapter 14 Probability

Access targeted multiple-choice questions for Chapter 14 Probability designed to align with the latest CBSE academic syllabus for Class 10 Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Chapter 14 Probability MCQs for Class 10 Mathematics

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: How many triangles can be formed by joining the vertices of a heptagon?
(a) 49
(b) 35
(c) 71
(d) 25
Show Answer & Explanation

Answer: (b) 35

Explanation:
1. Any 3 of the 7 vertices form a triangle: 7C3 = 35.

Teacher's Note:
1. Choose 3 vertices out of 7.
2. Use nCr.

Question 2: How many diagonals are there in an n-sided polygon (n > 3)?
(a) \( {}^nC_2 - n \)
(b) \( {}^nC_2 \)
(c) \( {}^{n(n-1)}C_2 \)
(d) \( {}^nC_{n-2} \)
Show Answer & Explanation

Answer: (a) \( {}^nC_2 - n \)

Explanation:
1. Joining any two of n vertices gives nC2 segments, of which n are sides.
2. Diagonals = nC2 - n.

Teacher's Note:
1. Count all segments first.
2. Subtract the sides.

Question 3: When two dice are thrown simultaneously, the probability that there is exactly one 5 is
(a) \( \frac{5}{18} \)
(b) \( \frac{1}{9} \)
(c) \( \frac{6}{23} \)
(d) \( \frac{7}{24} \)
Show Answer & Explanation

Answer: (a) \( \frac{5}{18} \)

Explanation:
1. Exactly one 5: (5, x) or (x, 5) with x ≠ 5: 5 + 5 = 10 outcomes.
2. P = 10/36 = 5/18.

Teacher's Note:
1. Count the outcomes in the 6 × 6 sample space.
2. Exclude (5, 5).

Question 4: When two fair dice are thrown, the probability that both show even numbers is
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{1}{5} \)
Show Answer & Explanation

Answer: (b) \( \frac{1}{4} \)

Explanation:
1. Even outcomes on each die: 3, so 3 × 3 = 9 favourable of 36.
2. P = 9/36 = 1/4.

Teacher's Note:
1. Each die has 3 even faces.
2. Multiply the counts.

Question 5: When 3 fair dice are thrown, the probability that the sum of the numbers is at least 6 is
(a) \( \frac{103}{216} \)
(b) \( \frac{47}{216} \)
(c) \( \frac{103}{108} \)
(d) \( \frac{1}{2} \)
Show Answer & Explanation

Answer: (c) \( \frac{103}{108} \)

Explanation:
1. Sum < 6 means sum 3, 4 or 5: 1 + 3 + 6 = 10 outcomes of 216.
2. P(sum ≥ 6) = 1 - 10/216 = 206/216 = 103/108.

Teacher's Note:
1. Use the complement.
2. Count ordered triples with sum 3, 4, 5.

Question 6: One coin has a tail and a head and the other has both heads. One of the two coins is chosen and tossed, and the result is a head. The probability that it is the coin with 2 heads is
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{2}{7} \)
(d) \( \frac{2}{3} \)
Show Answer & Explanation

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

Explanation:
1. The 3 equally likely faces showing heads: 1 on the first coin and 2 on the second.
2. P = 2/3.

Teacher's Note:
1. Count the head faces in all.
2. Two of the three belong to the double-head coin.

Question 7: Three coins are tossed. The probability of getting neither 3 heads nor 3 tails is
(a) \( \frac{3}{4} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{2}{3} \)
Show Answer & Explanation

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

Explanation:
1. Outcomes: 8. HHH and TTT are excluded, leaving 6.
2. P = 6/8 = 3/4.

Teacher's Note:
1. Use the complement.
2. Total outcomes = 8.

Question 8: If P(A) = 0.5, P(B) = 0.3 and P(A ∩ B) = 0.2, then P(A ∪ B) is
(a) \( \frac{2}{5} \)
(b) \( \frac{3}{5} \)
(c) \( \frac{3}{10} \)
(d) \( \frac{9}{10} \)
Show Answer & Explanation

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

Explanation:
1. P(A ∪ B) = 0.5 + 0.3 - 0.2 = 0.6 = 3/5.

Teacher's Note:
1. Use the addition theorem.
2. Subtract the overlap.

Question 9: A and B are mutually exclusive with P(A) = 1/2 and P(B) = 1/3. Then P(A ∪ B) is
(a) \( \frac{3}{4} \)
(b) \( \frac{2}{3} \)
(c) \( \frac{5}{6} \)
(d) \( \frac{1}{6} \)
Show Answer & Explanation

Answer: (c) \( \frac{5}{6} \)

Explanation:
1. P(A ∩ B) = 0, so P(A ∪ B) = 1/2 + 1/3 = 5/6.

Teacher's Note:
1. Mutually exclusive means no overlap.
2. Add the probabilities.

Question 10: In a throw of two dice, the probability that the sum is not 3 or 4 is
(a) \( \frac{4}{9} \)
(b) \( \frac{5}{36} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{31}{36} \)
Show Answer & Explanation

Answer: (d) \( \frac{31}{36} \)

Explanation:
1. Sum 3: 2 outcomes, sum 4: 3 outcomes, together 5.
2. P = 1 - 5/36 = 31/36.

Teacher's Note:
1. Use the complement.
2. Count outcomes with sum 3 and 4.

Question 11: Three identical dice are rolled. The probability that the same number appears on each of them is
(a) \( \frac{1}{36} \)
(b) \( \frac{1}{6} \)
(c) \( \frac{1}{216} \)
(d) \( \frac{3}{28} \)
Show Answer & Explanation

Answer: (a) \( \frac{1}{36} \)

Explanation:
1. There are 6 favourable outcomes (111, 222, ..., 666) of 216.
2. P = 6/216 = 1/36.

Teacher's Note:
1. Count triples with all equal numbers.
2. Total = 6³ = 216.

Question 12: Which of the following cannot be the probability of an event?
(a) \( \frac{2}{3} \)
(b) \( \frac{7}{5} \)
(c) 15%
(d) 0.33
Show Answer & Explanation

Answer: (b) \( \frac{7}{5} \)

Explanation:
1. A probability lies between 0 and 1 inclusive.
2. 7/5 > 1, so it is not possible; 2/3, 15% (0.15) and 0.33 are all valid.

Teacher's Note:
1. Check each value lies in [0, 1].
2. Convert the percentage.

Question 13: A card is drawn from a pack of 52 cards and a gambler bets that it is a spade or an ace. The odds against his winning are
(a) 4 : 9
(b) 7 : 6
(c) 9 : 4
(d) 8 : 5
Show Answer & Explanation

Answer: (c) 9 : 4

Explanation:
1. Favourable: 13 spades + 3 other aces = 16, so P(win) = 16/52 = 4/13.
2. Odds against = 36 : 16 = 9 : 4.

Teacher's Note:
1. Count spades and non-spade aces.
2. Odds against = failures : successes.

Question 14: Two dice are rolled. The probability of getting a total of 9 is
(a) \( \frac{1}{3} \)
(b) \( \frac{1}{9} \)
(c) \( \frac{8}{9} \)
(d) \( \frac{9}{10} \)
Show Answer & Explanation

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

Explanation:
1. Total 9: (3,6), (4,5), (5,4), (6,3), 4 outcomes.
2. P = 4/36 = 1/9.

Teacher's Note:
1. List pairs with sum 9.
2. Total outcomes = 36.

Question 15: Two dice are rolled. The probability that the sum is greater than 9 is
(a) \( \frac{10}{11} \)
(b) \( \frac{5}{6} \)
(c) \( \frac{1}{6} \)
(d) \( \frac{8}{9} \)
Show Answer & Explanation

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

Explanation:
1. Sum 10, 11, 12: 3 + 2 + 1 = 6 outcomes.
2. P = 6/36 = 1/6.

Teacher's Note:
1. Count sums 10, 11 and 12.
2. Divide by 36.

Question 16: Two dice are rolled. The probability of a total of 9 or 11 is
(a) \( \frac{2}{99} \)
(b) \( \frac{20}{99} \)
(c) \( \frac{1}{10} \)
(d) \( \frac{1}{6} \)
Show Answer & Explanation

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

Explanation:
1. Total 9: 4 outcomes, total 11: 2 outcomes.
2. P = 6/36 = 1/6.

Teacher's Note:
1. The events are mutually exclusive.
2. Add the counts.

Question 17: Two dice are rolled. The probability of getting a doublet is
(a) \( \frac{1}{6} \)
(b) \( \frac{1}{12} \)
(c) \( 0 \)
(d) \( \frac{5}{8} \)
Show Answer & Explanation

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

Explanation:
1. Doublets: (1,1), ..., (6,6): 6 outcomes.
2. P = 6/36 = 1/6.

Teacher's Note:
1. Doublet means both numbers equal.
2. Total = 36.

Question 18: Two dice are rolled. The probability of a doublet of even numbers is
(a) \( \frac{5}{8} \)
(b) \( \frac{1}{12} \)
(c) \( \frac{3}{4} \)
(d) \( \frac{1}{4} \)
Show Answer & Explanation

Answer: (b) \( \frac{1}{12} \)

Explanation:
1. Even doublets: (2,2), (4,4), (6,6): 3 outcomes.
2. P = 3/36 = 1/12.

Teacher's Note:
1. List even doublets.
2. Divide by 36.

Question 19: Two dice are rolled. The probability of a multiple of 2 on one die and a multiple of 3 on the other is
(a) \( \frac{5}{12} \)
(b) \( \frac{25}{36} \)
(c) \( \frac{11}{36} \)
(d) \( \frac{5}{6} \)
Show Answer & Explanation

Answer: (c) \( \frac{11}{36} \)

Explanation:
1. Pairs with first die even and second in {3, 6}: 6. Pairs with first in {3, 6} and second even: 6. Only (6, 6) is in both lists.
2. Total distinct = 6 + 6 - 1 = 11.
3. P = 11/36.

Teacher's Note:
1. Count by inclusion-exclusion.
2. The pair (6, 6) satisfies both descriptions.

Question 20: Two dice are rolled. The probability that the sum is divisible by 3 or 4 is
(a) \( \frac{2}{3} \)
(b) \( \frac{1}{7} \)
(c) \( \frac{7}{12} \)
(d) \( \frac{5}{9} \)
Show Answer & Explanation

Answer: (d) \( \frac{5}{9} \)

Explanation:
1. Sum divisible by 3: 3, 6, 9, 12: 2 + 5 + 4 + 1 = 12 outcomes. Divisible by 4: 4, 8, 12: 3 + 5 + 1 = 9 outcomes. Both (sum 12): 1.
2. Favourable = 12 + 9 - 1 = 20, so P = 20/36 = 5/9.

Teacher's Note:
1. Use inclusion-exclusion.
2. The sum 12 is in both lists.

Question 21: Two dice are rolled. The probability that the sum is a prime number is
(a) \( \frac{5}{12} \)
(b) \( \frac{3}{5} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{3}{4} \)
Show Answer & Explanation

Answer: (a) \( \frac{5}{12} \)

Explanation:
1. Prime sums 2, 3, 5, 7, 11: 1 + 2 + 4 + 6 + 2 = 15 outcomes.
2. P = 15/36 = 5/12.

Teacher's Note:
1. List prime sums from 2 to 12.
2. Count each.

Question 22: Two dice are rolled. The probability of getting at least one 5 is
(a) \( \frac{3}{5} \)
(b) \( \frac{11}{36} \)
(c) \( \frac{1}{5} \)
(d) \( \frac{5}{36} \)
Show Answer & Explanation

Answer: (b) \( \frac{11}{36} \)

Explanation:
1. Complement: no 5 on either die: 5 × 5 = 25 outcomes.
2. P = 1 - 25/36 = 11/36.

Teacher's Note:
1. Use the complement.
2. Total = 36.

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Chapter 14 Probability Objective Questions & Solutions for Class 10 Mathematics

About Chapter 14 Probability MCQs for Class 10 Mathematics

Test your conceptual understanding of Chapter 14 Probability with these targeted multiple-choice questions. Designed in alignment with the latest CBSE curriculum for Class 10 Mathematics, these problem sets build accuracy and prepare students for objective exams.

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FAQs

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