CBSE Class 10 Mathematics Probability MCQs Set 16

Download CBSE 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.

Chapter-wise Objective Questions: Chapter 14 Probability

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: Three unbiased coins are tossed. The probability of getting at most two heads is:
(a) \( \frac{7}{8} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{3}{4} \)
(d) \( \frac{1}{8} \)
Show Answer & Explanation

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

Explanation:
1. The only outcome excluded is HHH. P = 1 − 1/8 = 7/8.

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Question 2: Two dice are thrown together. The probability that the sum is at least 9 is:
(a) \( \frac{1}{6} \)
(b) \( \frac{4}{9} \)
(c) \( \frac{7}{36} \)
(d) \( \frac{5}{18} \)
Show Answer & Explanation

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

Explanation:
1. Sums 9, 10, 11, 12 occur 4 + 3 + 2 + 1 = 10 ways of 36, so P = 5/18.

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Question 3: A letter is drawn at random from the word ERROR. The probability of drawing the letter R is:
(a) \( \frac{1}{5} \)
(b) \( \frac{2}{5} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{3}{5} \)
Show Answer & Explanation

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

Explanation:
1. ERROR has 5 letters: E, R, R, O, R. R occurs 3 times, so P = 3/5.

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Question 4: A card is drawn from a pack of 52. The probability that it is either black or an ace is:
(a) \( \frac{1}{2} \)
(b) \( \frac{15}{26} \)
(c) \( \frac{7}{13} \)
(d) \( \frac{27}{52} \)
Show Answer & Explanation

Answer: (c) \( \frac{7}{13} \)

Explanation:
1. Black: 26, aces: 4, black aces: 2. Black or ace = 28, so P = 28/52 = 7/13.

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Question 5: The number of triangles that can be formed by joining the vertices of a heptagon is:
(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.

Question 6: Mr and Mrs X live with their seven children, and the female to male ratio in the family (9 members) is 1 : 2. The probability that all the children are of the same sex is:
(a) \( 0 \)
(b) \( \frac{2}{5} \)
(c) \( \frac{1}{21} \)
(d) \( \frac{1}{42} \)
Show Answer & Explanation

Answer: (a) \( 0 \)

Explanation:
1. The family has 3 females and 6 males. With the mother, there are 2 girls and 5 boys among the children. Not all children can be of the same sex, so P = 0.

Teacher's Note:
1. Count the girls.

Question 7: In the same family (seven children: 5 boys and 2 girls, in random order of birth), the probability that the five elder children are boys and the youngest two are girls is:
(a) \( \frac{1}{21} \)
(b) \( 0 \)
(c) \( \frac{2}{5} \)
(d) \( \frac{1}{42} \)
Show Answer & Explanation

Answer: (a) \( \frac{1}{21} \)

Explanation:
1. The positions of the 2 girls can be chosen in 7C2 = 21 equally likely ways. Only one of these puts them at the youngest two places. P = 1/21.

Teacher's Note:
1. Use combinations for the positions of the girls.

Question 8: When two dice are thrown, the probability that exactly one die shows a 5 is:
(a) \( \frac{1}{9} \)
(b) \( \frac{11}{36} \)
(c) \( \frac{1}{6} \)
(d) \( \frac{5}{18} \)
Show Answer & Explanation

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

Explanation:
1. Exactly one 5: 5 on the first die and non-5 on the second (5 ways) plus the reverse (5 ways) = 10 of 36. P = 5/18.

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Question 9: When two fair dice are thrown, the probability that both dice show even numbers is:
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{1}{4} \)
(d) \( \frac{1}{5} \)
Show Answer & Explanation

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

Explanation:
1. 3 × 3 = 9 of 36, so P = 1/4.

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Question 10: A card is drawn from a pack of 52. The probability that the card is a queen is:
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{13} \)
(c) \( \frac{1}{26} \)
(d) \( \frac{7}{13} \)
Show Answer & Explanation

Answer: (b) \( \frac{1}{13} \)

Explanation:
1. There are 4 queens in 52 cards: 4/52 = 1/13.

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Question 11: A card is drawn from a pack of 52. The probability that the card is black and a queen is:
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{26} \)
(c) \( \frac{1}{13} \)
(d) \( \frac{7}{13} \)
Show Answer & Explanation

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

Explanation:
1. The black queens are the queen of spades and the queen of clubs: 2/52 = 1/26.

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Question 12: 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 HHH and TTT are the only two excluded: 6 of 8 remain, so P = 3/4.

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Question 13: A and B are two events with P(A) = 0.5, P(B) = 0.3 and P(A and B) = 0.2. The probability P(A or B) is:
(a) \( \frac{2}{5} \)
(b) \( \frac{3}{10} \)
(c) \( \frac{9}{10} \)
(d) \( \frac{3}{5} \)
Show Answer & Explanation

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

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

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Question 14: Two dice are thrown. The probability that the sum is neither 3 nor 4 is:
(a) \( \frac{4}{9} \)
(b) \( \frac{5}{36} \)
(c) \( \frac{31}{36} \)
(d) \( \frac{1}{3} \)
Show Answer & Explanation

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

Explanation:
1. Sum 3: 2 ways; sum 4: 3 ways; so 5 outcomes are excluded. P = 31/36.

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Question 15: Three identical dice are rolled. The probability that the same number appears on each is:
(a) \( \frac{1}{6} \)
(b) \( \frac{1}{216} \)
(c) \( \frac{1}{36} \)
(d) \( \frac{1}{18} \)
Show Answer & Explanation

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

Explanation:
1. There are 6 outcomes with all faces equal out of 216. P = 6/216 = 1/36.

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Question 16: Which of the following cannot be the probability of an event?
(a) \( \frac23 \)
(b) \( \frac75 \)
(c) \( \frac12 \)
(d) 15%
Show Answer & Explanation

Answer: (b) \( \frac75 \)

Explanation:
1. A probability lies between 0 and 1. 7/5 is greater than 1, while 2/3, 1/2 and 15% = 0.15 are all within [0, 1].

Teacher's Note:
1. 0 ≤ P ≤ 1.

Question 17: A card is drawn from an ordinary pack of 52 and a gambler bets that it is a spade or an ace. The odds against his winning are:
(a) 9 : 4
(b) 4 : 9
(c) 7 : 6
(d) 8 : 5
Show Answer & Explanation

Answer: (a) 9 : 4

Explanation:
1. Spade or ace: 13 + 4 − 1 = 16 cards. P(win) = 16/52 = 4/13 and P(lose) = 9/13. Odds against = 9 : 4.

Teacher's Note:
1. Odds against = P(not E) : P(E).

Question 18: Two dice are rolled. The probability of getting a sum greater than 9 is:
(a) \( \frac{1}{3} \)
(b) \( \frac{5}{6} \)
(c) \( \frac{10}{11} \)
(d) \( \frac{1}{6} \)
Show Answer & Explanation

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

Explanation:
1. Sums 10, 11, 12: 3 + 2 + 1 = 6 of 36, so P = 1/6.

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Question 19: Two dice are rolled. The probability of getting 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 ways; total 11: 2 ways. 6 of 36 gives P = 1/6.

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Question 20: Two dice are rolled. The probability of getting a doublet of even numbers is:
(a) \( \frac{5}{8} \)
(b) \( \frac{3}{4} \)
(c) \( \frac{1}{12} \)
(d) \( \frac{1}{4} \)
Show Answer & Explanation

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

Explanation:
1. The doublets (2,2), (4,4), (6,6): 3 of 36, so P = 1/12.

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Question 21: Two dice are rolled. The probability that the sum of the numbers is divisible by 3 or 4 is:
(a) \( \frac{2}{3} \)
(b) \( \frac{5}{9} \)
(c) \( \frac{1}{7} \)
(d) \( \frac{7}{12} \)
Show Answer & Explanation

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

Explanation:
1. Sums divisible by 3: 3, 6, 9, 12 (2 + 5 + 4 + 1 = 12 ways). Sums divisible by 4: 4, 8, 12 (3 + 5 + 1 = 9 ways). The sum 12 is in both (1 way). 12 + 9 − 1 = 20 of 36, so P = 5/9.

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Question 22: 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 (1 way), 3 (2), 5 (4), 7 (6), 11 (2) = 15 of 36, so P = 5/12.

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Question 23: Two dice are rolled. The probability of getting at least one 5 is:
(a) \( \frac{11}{36} \)
(b) \( \frac{3}{5} \)
(c) \( \frac{1}{5} \)
(d) \( \frac{5}{36} \)
Show Answer & Explanation

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

Explanation:
1. No 5 at all: 5 × 5 = 25 outcomes. P = 1 − 25/36 = 11/36.

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Question 24: Three fair coins are tossed. The probability of getting exactly one head is:
(a) \( \frac{1}{8} \)
(b) \( \frac{7}{8} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{3}{8} \)
Show Answer & Explanation

Answer: (d) \( \frac{3}{8} \)

Explanation:
1. Outcomes HTT, THT, TTH: 3 of 8, so P = 3/8 = 0.375.

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Question 25: Three fair coins are tossed. The probability of getting exactly two tails is:
(a) \( \frac{1}{8} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{3}{8} \)
(d) \( \frac{7}{8} \)
Show Answer & Explanation

Answer: (c) \( \frac{3}{8} \)

Explanation:
1. Exactly two tails means one head: HTT, THT, TTH are 3 of 8 outcomes, so P = 3/8.

Teacher's Note:
1. P(E) = favourable outcomes / total outcomes.

Free study material for Mathematics

Practice MCQs for Class 10 Mathematics Chapter 14 Probability

About Chapter 14 Probability MCQs for Class 10 Mathematics

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

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Each question includes structured solution keys mapped directly to standard CBSE textbooks, helping students evaluate their reasoning and correct mistakes early in their revision.

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FAQs

Where can I access latest CBSE Class 10 Mathematics Probability MCQs Set 16?

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Are Assertion-Reasoning and Case-Study MCQs included in the Mathematics Class 10 material?

Yes, our CBSE Class 10 Mathematics Probability MCQs Set 16 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the CBSE paper is now competency-based.

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By solving our CBSE Class 10 Mathematics Probability MCQs Set 16, Class 10 students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Mathematics.

Do you provide answers and explanations for CBSE Class 10 Mathematics Probability MCQs Set 16?

Yes, Mathematics MCQs for Class 10 have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused CBSE exams.

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