CBSE Class 10 Mathematics Probability MCQs Set 15

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: Three dice are thrown together. The probability that the total is at least 6 is:
(a) \( \frac{103}{108} \)
(b) \( \frac{103}{216} \)
(c) \( \frac{5}{108} \)
(d) \( \frac{47}{216} \)
Show Answer & Explanation

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

Explanation:
1. There are 216 outcomes. Totals less than 6 are 3, 4, 5: (1,1,1), three with total 4 and six with total 5 make 10. P = 1 − 10/216 = 103/108.

Teacher's Note:
1. Use the complement.

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

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

Explanation:
1. The primes are 2, 3, 5, so P = 3/6 = 1/2.

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

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

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

Explanation:
1. Even: {2, 4, 6}; multiples of 3: {3, 6}. The union is {2, 3, 4, 6}, so P = 4/6 = 2/3.

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

Question 4: A card is drawn at random from a well-shuffled pack of 52 cards. The probability that it is a black face card is:
(a) \( \frac{3}{13} \)
(b) \( \frac{3}{26} \)
(c) \( \frac{1}{13} \)
(d) \( \frac{6}{13} \)
Show Answer & Explanation

Answer: (b) \( \frac{3}{26} \)

Explanation:
1. Black face cards: jack, queen, king of spades and clubs = 6. P = 6/52 = 3/26.

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

Question 5: In a non-leap year, the probability of getting 53 Tuesdays is:
(a) \( \frac{2}{7} \)
(b) \( \frac{1}{7} \)
(c) \( \frac{1}{52} \)
(d) \( \frac{53}{365} \)
Show Answer & Explanation

Answer: (b) \( \frac{1}{7} \)

Explanation:
1. The year has 52 full weeks and 1 extra day, which can be any of the 7 days with equal chance. P(Tuesday) = 1/7.

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

Question 6: A bag contains 9 red, 7 white and 4 black balls. The probability that a ball drawn is red is:
(a) \( \frac{9}{20} \)
(b) \( \frac{11}{20} \)
(c) \( \frac{7}{20} \)
(d) \( \frac{1}{5} \)
Show Answer & Explanation

Answer: (a) \( \frac{9}{20} \)

Explanation:
1. 9 of the 20 balls are red, so P = 9/20.

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

Question 7: The probability that a leap year selected at random contains 53 Sundays is:
(a) \( \frac{1}{7} \)
(b) \( \frac{3}{7} \)
(c) \( \frac{1}{366} \)
(d) \( \frac{2}{7} \)
Show Answer & Explanation

Answer: (d) \( \frac{2}{7} \)

Explanation:
1. A leap year has 52 weeks plus 2 extra days. The pairs of extra days are (Mon,Tue), (Tue,Wed), ..., (Sun,Mon): 7 pairs, of which 2 contain Sunday. P = 2/7.

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

Question 8: A card is drawn from a pack of 52. The probability that it is either red or a king is:
(a) \( \frac{1}{2} \)
(b) \( \frac{15}{26} \)
(c) \( \frac{7}{13} \)
(d) \( \frac{3}{13} \)
Show Answer & Explanation

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

Explanation:
1. Red: 26, kings: 4, red kings: 2. Red or king = 26 + 4 − 2 = 28. P = 28/52 = 7/13.

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

Question 9: A card is drawn from a pack of 52. The probability that it is neither a heart nor a king is:
(a) \( \frac{10}{13} \)
(b) \( \frac{36}{13} \)
(c) \( \frac{9}{13} \)
(d) \( \frac{3}{4} \)
Show Answer & Explanation

Answer: (c) \( \frac{9}{13} \)

Explanation:
1. Heart or king = 13 + 4 − 1 = 16. Neither = 52 − 16 = 36. P = 36/52 = 9/13.

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

Question 10: In a single throw of two dice, the probability of a total of 9 is:
(a) \( \frac{1}{6} \)
(b) \( \frac{1}{9} \)
(c) \( \frac{1}{12} \)
(d) \( \frac{5}{36} \)
Show Answer & Explanation

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

Explanation:
1. The pairs (3,6), (4,5), (5,4), (6,3) give 4 outcomes of 36. P = 1/9.

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

Question 11: In a single throw of two dice, the probability of getting a doublet is:
(a) \( \frac{1}{6} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{1}{9} \)
(d) \( \frac{1}{3} \)
Show Answer & Explanation

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

Explanation:
1. The doublets (1,1) to (6,6) are 6 of 36 outcomes. P = 1/6.

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

Question 12: In a single throw of two dice, the probability of getting a multiple of 2 on one die and a multiple of 3 on the other is:
(a) \( \frac{1}{3} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{5}{18} \)
(d) \( \frac{11}{36} \)
Show Answer & Explanation

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

Explanation:
1. Even on the first and a multiple of 3 on the second: 3 × 2 = 6. The reverse: 6. Both counted twice: (6,6) appears in both, so 6 + 6 − 1 = 11 of 36.

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

Question 13: In a single throw of two dice, the probability of an odd number on one die and 6 on the other is:
(a) \( \frac{1}{12} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{1}{6} \)
Show Answer & Explanation

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

Explanation:
1. Odd on the first and 6 on the second: 3. The reverse: 3. Total 6 of 36 (no overlap), so P = 1/6.

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

Question 14: In a single throw of two dice, the probability of a total of 11 is:
(a) \( \frac{1}{36} \)
(b) \( \frac{1}{12} \)
(c) \( \frac{1}{18} \)
(d) \( \frac{1}{9} \)
Show Answer & Explanation

Answer: (c) \( \frac{1}{18} \)

Explanation:
1. The outcomes (5,6) and (6,5) give 2 of 36, so P = 1/18.

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

Question 15: Two coins are tossed together. The probability of getting exactly one head is:
(a) \( \frac{1}{4} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{3}{4} \)
(d) \( \frac{1}{3} \)
Show Answer & Explanation

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

Explanation:
1. Outcomes HT and TH are 2 of 4. P = 1/2.

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

Question 16: A die is thrown twice. The probability that both throws show a prime number is:
(a) \( \frac{1}{4} \)
(b) \( \frac{1}{6} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{1}{9} \)
Show Answer & Explanation

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

Explanation:
1. Primes: 2, 3, 5. 3 × 3 = 9 of 36, so P = 1/4.

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

Question 17: Three unbiased coins are tossed. The probability of exactly two heads is:
(a) \( \frac{3}{8} \)
(b) \( \frac{1}{8} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{1}{4} \)
Show Answer & Explanation

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

Explanation:
1. HHT, HTH, THH: 3 of 8 outcomes, so P = 3/8.

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

Question 18: One card is drawn from a pack of 52. The probability that it is a jack, queen, king or ace is:
(a) \( \frac{1}{13} \)
(b) \( \frac{2}{13} \)
(c) \( \frac{8}{13} \)
(d) \( \frac{4}{13} \)
Show Answer & Explanation

Answer: (d) \( \frac{4}{13} \)

Explanation:
1. There are 4 × 4 = 16 such cards. P = 16/52 = 4/13.

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

Question 19: A bag contains 3 red, 5 black and 7 white balls. The probability that a ball drawn is not black is:
(a) \( \frac{1}{3} \)
(b) \( \frac{7}{15} \)
(c) \( \frac{2}{3} \)
(d) \( \frac{1}{5} \)
Show Answer & Explanation

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

Explanation:
1. Not black: 3 + 7 = 10 of 15. P = 2/3.

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

Question 20: The probability that two people chosen at random have the same birthday (365 days in the year) is:
(a) \( \frac{2}{365} \)
(b) \( \frac{1}{365} \)
(c) \( \frac{1}{730} \)
(d) \( \frac{1}{100000} \)
Show Answer & Explanation

Answer: (b) \( \frac{1}{365} \)

Explanation:
1. The first person's birthday can be any day; the second matches it with probability 1/365.

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

Question 21: Two dice are tossed together. The probability of a doublet or a total of 6 is:
(a) \( \frac{1}{6} \)
(b) \( \frac{5}{18} \)
(c) \( \frac{11}{36} \)
(d) \( \frac{7}{36} \)
Show Answer & Explanation

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

Explanation:
1. Doublets: 6; total 6: (1,5), (2,4), (3,3), (4,2), (5,1) = 5; (3,3) is in both. 6 + 5 − 1 = 10 of 36, so P = 5/18.

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

Question 22: A natural number is chosen at random from the first 500. The probability that it is divisible by 3 or 5 is:
(a) \( \frac{233}{500} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{133}{250} \)
(d) \( \frac{83}{250} \)
Show Answer & Explanation

Answer: (a) \( \frac{233}{500} \)

Explanation:
1. Multiples of 3: 166; of 5: 100; of 15: 33. Either: 166 + 100 − 33 = 233. P = 233/500.

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

Question 23: A card is drawn at random from a well-shuffled deck of 52 cards. The probability that it is a spade or a king is:
(a) \( \frac{1}{4} \)
(b) \( \frac{17}{52} \)
(c) \( \frac{3}{13} \)
(d) \( \frac{4}{13} \)
Show Answer & Explanation

Answer: (d) \( \frac{4}{13} \)

Explanation:
1. Spades 13, kings 4, king of spades counted twice: 13 + 4 − 1 = 16. P = 16/52 = 4/13.

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

Question 24: A die is thrown. The probability that the number shown is less than or equal to 4 is:
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{2}{3} \)
(d) \( \frac{5}{6} \)
Show Answer & Explanation

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

Explanation:
1. The outcomes 1, 2, 3, 4 give 4 of 6, so P = 2/3.

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

Question 25: A die is thrown. The probability of a number greater than 6 is:
(a) \( \frac{1}{6} \)
(b) \( \frac{1}{2} \)
(c) \( 0 \)
(d) \( 1 \)
Show Answer & Explanation

Answer: (c) \( 0 \)

Explanation:
1. No face shows more than 6, so the event is impossible and P = 0.

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

Question 26: Three unbiased coins are tossed. The probability of getting at least two heads is:
(a) \( \frac{3}{8} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{1}{4} \)
(d) \( \frac{7}{8} \)
Show Answer & Explanation

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

Explanation:
1. Two heads: 3 outcomes; three heads: 1. Total 4 of 8, so P = 1/2.

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

Free study material for Mathematics

Practice MCQs for Class 10 Mathematics Chapter 14 Probability

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

Explore reliable practice questions for Chapter 14 Probability tailored for Class 10 Mathematics learners. Use these multiple-choice formats to evaluate preparedness and strengthen problem-solving skills.

Concept Clarification for Chapter 14 Probability

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

Additional Study Resources for Class 10 Mathematics

Follow up your worksheet practice by attempting the interactive online Mathematics MCQ test for this chapter to evaluate your execution speed. All platform resources are free to access.

FAQs

Where can I access latest CBSE Class 10 Mathematics Probability MCQs Set 15?

You can get most exhaustive CBSE Class 10 Mathematics Probability MCQs Set 15 for free on StudiesToday.com. These MCQs for Class 10 Mathematics are updated for the 2026-27 academic session as per CBSE examination standards.

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

Yes, our CBSE Class 10 Mathematics Probability MCQs Set 15 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the CBSE paper is now competency-based.

How do practicing Mathematics MCQs help in scoring full marks in Class 10 exams?

By solving our CBSE Class 10 Mathematics Probability MCQs Set 15, 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 15?

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.

Can I practice these Mathematics Class 10 MCQs online?

Yes, you can also access online interactive tests for CBSE Class 10 Mathematics Probability MCQs Set 15 on StudiesToday.com as they provide instant answers and score to help you track your progress in Mathematics.