CBSE Class 10 Mathematics Probability MCQs Set 19

Practice 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.

Access Chapter 14 Probability Questions and Solutions

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: A box contains 5 red, 4 green and 7 white balls. One ball is drawn at random. The probability that it is white is:
(a) \( \frac{7}{16} \)
(b) \( \frac{9}{16} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{3}{8} \)
Show Answer & Explanation

Answer: (a) \( \frac{7}{16} \)

Explanation:
1. Total balls = 16 and white balls = 7. P = 7/16.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 2: A box contains 5 red, 4 green and 7 white balls. One ball is drawn at random. The probability that it is neither red nor white is:
(a) \( \frac{3}{4} \)
(b) \( \frac{5}{16} \)
(c) \( \frac{3}{16} \)
(d) \( \frac{1}{4} \)
Show Answer & Explanation

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

Explanation:
1. Neither red nor white means green: 4 balls. P = 4/16 = 1/4.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 3: All three face cards of spades are removed from a pack of 52 cards and one card is drawn from the remaining cards. The probability that it is a black face card is:
(a) \( \frac{46}{49} \)
(b) \( \frac{4}{49} \)
(c) \( \frac{3}{49} \)
(d) \( \frac{2}{49} \)
Show Answer & Explanation

Answer: (c) \( \frac{3}{49} \)

Explanation:
1. 49 cards remain. Black face cards: 6 - 3 = 3 (the face cards of clubs). P = 3/49.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 4: All three face cards of spades are removed from a pack of 52 cards and one card is drawn from the remaining cards. The probability that it is a queen is:
(a) \( \frac{46}{49} \)
(b) \( \frac{3}{49} \)
(c) \( \frac{4}{49} \)
(d) \( \frac{2}{49} \)
Show Answer & Explanation

Answer: (b) \( \frac{3}{49} \)

Explanation:
1. The queen of spades is removed, so 3 queens remain among 49 cards. P = 3/49.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 5: All three face cards of spades are removed from a pack of 52 cards and one card is drawn from the remaining cards. The probability that it is a black card is:
(a) \( \frac{26}{49} \)
(b) \( \frac{23}{49} \)
(c) \( \frac{24}{49} \)
(d) \( \frac{22}{49} \)
Show Answer & Explanation

Answer: (b) \( \frac{23}{49} \)

Explanation:
1. Black cards: 26 - 3 = 23 among 49. P = 23/49.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 6: A die is thrown once. The probability of getting a prime number is:
(a) \( \frac{1}{2} \)
(b) \( \frac{2}{3} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{3}{7} \)
Show Answer & Explanation

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

Explanation:
1. The primes on a die are 2, 3, 5. P = 3/6 = 1/2.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 7: A die is thrown once. The probability of getting a multiple of 2 or 3 is:
(a) \( \frac{1}{3} \)
(b) \( \frac{5}{6} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{2}{3} \)
Show Answer & Explanation

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

Explanation:
1. The numbers are 2, 3, 4, 6. P = 4/6 = 2/3.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 8: A die is thrown once. The probability of getting a number greater than 3 is:
(a) \( \frac{2}{3} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{3}{7} \)
Show Answer & Explanation

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

Explanation:
1. The numbers are 4, 5, 6. P = 3/6 = 1/2.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 9: Two unbiased coins are tossed together. The probability of getting at least one head is:
(a) \( \frac{1}{4} \)
(b) \( 1 \)
(c) \( \frac{3}{4} \)
(d) \( \frac{1}{2} \)
Show Answer & Explanation

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

Explanation:
1. The outcomes HH, HT, TH, TT; at least one head in HH, HT, TH. P = 3/4.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 10: Two unbiased coins are tossed together. The probability of getting at most one head is:
(a) \( \frac{1}{4} \)
(b) \( \frac{3}{4} \)
(c) \( 1 \)
(d) \( \frac{1}{2} \)
Show Answer & Explanation

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

Explanation:
1. At most one head: HT, TH, TT. P = 3/4.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 11: A box contains 20 balls numbered 1 to 20. One ball is drawn at random. The probability that the number is divisible by 2 or 3 is:
(a) \( \frac{13}{20} \)
(b) \( \frac{7}{20} \)
(c) \( \frac{7}{10} \)
(d) \( \frac{3}{5} \)
Show Answer & Explanation

Answer: (a) \( \frac{13}{20} \)

Explanation:
1. The numbers are 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20: 13 numbers. P = 13/20.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 12: A box contains 20 balls numbered 1 to 20. One ball is drawn at random. The probability that the number is prime is:
(a) \( \frac{3}{5} \)
(b) \( \frac{9}{20} \)
(c) \( \frac{7}{20} \)
(d) \( \frac{2}{5} \)
Show Answer & Explanation

Answer: (d) \( \frac{2}{5} \)

Explanation:
1. The primes are 2, 3, 5, 7, 11, 13, 17, 19: 8 numbers. P = 8/20 = 2/5.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 13: A small die is dropped at random on a rectangular region 3 m by 2 m. The probability that it lands inside a circle of diameter 1 m drawn in the region is:
(a) \( \frac{\pi}{12} \)
(b) \( \frac{\pi}{6} \)
(c) \( \frac{\pi}{3} \)
(d) \( \frac{\pi}{24} \)
Show Answer & Explanation

Answer: (d) \( \frac{\pi}{24} \)

Explanation:
1. Area of the region = 6 m². Area of the circle = π(0.5)² = π/4 m². P = (π/4)/6 = π/24.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 14: Three coins are tossed together. The probability of getting at least two heads is:
(a) \( \frac{5}{8} \)
(b) \( \frac{3}{8} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{4}{9} \)
Show Answer & Explanation

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

Explanation:
1. Outcomes with 2 or 3 heads: HHT, HTH, THH, HHH = 4 of 8. P = 1/2.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 15: A bag contains 3 green, 4 blue and 2 orange marbles. One marble is picked at random. The probability that it is not orange is:
(a) \( \frac{2}{9} \)
(b) \( \frac{7}{9} \)
(c) \( \frac{8}{9} \)
(d) \( \frac{2}{3} \)
Show Answer & Explanation

Answer: (b) \( \frac{7}{9} \)

Explanation:
1. Total = 9 and 7 are not orange. P = 7/9.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 16: A number is selected at random from 1 to 27. The probability that it is prime is:
(a) \( \frac{1}{3} \)
(b) \( \frac{2}{3} \)
(c) \( \frac{10}{27} \)
(d) \( \frac{8}{27} \)
Show Answer & Explanation

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

Explanation:
1. The primes are 2, 3, 5, 7, 11, 13, 17, 19, 23: 9 numbers. P = 9/27 = 1/3.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 17: If P(E) = 0.05, then P(not E) is:
(a) 0.95
(b) -0.05
(c) 0.5
(d) 0.9
Show Answer & Explanation

Answer: (a) 0.95

Explanation:
1. P(not E) = 1 - 0.05 = 0.95.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 18: A bulb is taken at random from a box of 600 bulbs of which 12 are defective. The probability that it is non-defective is:
(a) 0.02
(b) 0.50
(c) 0.88
(d) 0.98
Show Answer & Explanation

Answer: (d) 0.98

Explanation:
1. Non-defective bulbs = 588. P = 588/600 = 0.98.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 19: Two dice are thrown together. The probability that the sum is even is:
(a) \( \frac{19}{36} \)
(b) \( \frac{17}{36} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{18}{37} \)
Show Answer & Explanation

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

Explanation:
1. A sum is even when both numbers have the same parity: 18 of the 36 outcomes. P = 1/2.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 20: Two dice are thrown together. The probability that the sum is a prime number is:
(a) \( \frac{7}{12} \)
(b) \( \frac{5}{12} \)
(c) \( \frac{4}{9} \)
(d) \( \frac{7}{18} \)
Show Answer & Explanation

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

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

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 21: Two dice are thrown together. The probability of a total of at least 10 is:
(a) \( \frac{5}{6} \)
(b) \( \frac{1}{6} \)
(c) \( \frac{7}{36} \)
(d) \( \frac{5}{36} \)
Show Answer & Explanation

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

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

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 22: Two dice are thrown together. The probability of getting a multiple of 2 on one die and a multiple of 3 on the other is:
(a) \( \frac{11}{36} \)
(b) \( \frac{25}{36} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{5}{18} \)
Show Answer & Explanation

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

Explanation:
1. Ordered pairs (first, second) with the first even and the second a multiple of 3: 3 x 2 = 6. With the second even and the first a multiple of 3: another 6. The pair (6, 6) is in both lists, so there are 6 + 6 - 1 = 11 outcomes. P = 11/36.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 23: The probability that a leap year chosen at random has 53 Tuesdays is:
(a) \( \frac{5}{7} \)
(b) \( \frac{3}{7} \)
(c) \( \frac{1}{7} \)
(d) \( \frac{2}{7} \)
Show Answer & Explanation

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

Explanation:
1. A leap year has 52 weeks and 2 extra days. The pair of extra days is one of 7 equally likely pairs, and 2 of them (Mon-Tue, Tue-Wed) include a Tuesday. P = 2/7.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 24: In a class of 18 girls and 16 boys, a monitor is chosen by drawing a name from a basket. The probability that a girl is chosen is:
(a) \( \frac{8}{17} \)
(b) \( \frac{19}{34} \)
(c) \( \frac{9}{17} \)
(d) \( \frac{1}{2} \)
Show Answer & Explanation

Answer: (c) \( \frac{9}{17} \)

Explanation:
1. Total pupils = 34. P = 18/34 = 9/17.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 25: A jar contains only green, white and yellow marbles. The probability of a green marble is 1/4 and of a white marble is 1/3. If the jar has 10 yellow marbles, the total number of marbles is:
(a) 12
(b) 30
(c) 24
(d) 40
Show Answer & Explanation

Answer: (c) 24

Explanation:
1. P(yellow) = 1 - 1/4 - 1/3 = 5/12. So 10/N = 5/12 and N = 24.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 26: A card is drawn from a well-shuffled pack of 52. The probability that it is a spade or an ace is:
(a) \( \frac{9}{13} \)
(b) \( \frac{4}{13} \)
(c) \( \frac{17}{52} \)
(d) \( \frac{15}{52} \)
Show Answer & Explanation

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

Explanation:
1. Spades 13 plus 3 aces of other suits = 16. P = 16/52 = 4/13.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

Question 27: A card is drawn from a well-shuffled pack of 52. The probability that it is a red king is:
(a) \( \frac{1}{26} \)
(b) \( \frac{25}{26} \)
(c) \( \frac{3}{52} \)
(d) \( \frac{1}{52} \)
Show Answer & Explanation

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

Explanation:
1. Two red kings. P = 2/52 = 1/26.

Teacher's Note:
1. P(E) = number of favourable outcomes / total number of equally likely outcomes. Always 0 ≤ P(E) ≤ 1, and P(not E) = 1 - P(E).
2. List the sample space carefully before counting favourable outcomes.

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Practice MCQs for Class 10 Mathematics Chapter 14 Probability

About Chapter 14 Probability MCQs for Class 10 Mathematics

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FAQs

Where can I access latest CBSE Class 10 Mathematics Probability MCQs Set 19?

You can get most exhaustive CBSE Class 10 Mathematics Probability MCQs Set 19 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 19 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 19, 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 19?

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 19 on StudiesToday.com as they provide instant answers and score to help you track your progress in Mathematics.