CBSE Class 10 Mathematics Probability MCQs Set 18

Mathematics Objective Questions and Answers: Chapter 14 Probability

Explore reliable objective questions for Chapter 14 Probability tailored for Class 10 learners. Utilizing these Mathematics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Download Chapter 14 Probability MCQs with Answers

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Multiple Choice Questions

Question 1: A card is drawn from 17 cards numbered 1, 2, ..., 17. The probability that the number is a multiple of 3 or 7 is:
(a) \( \frac{7}{17} \)
(b) \( \frac{10}{17} \)
(c) \( \frac{8}{17} \)
(d) \( \frac{6}{17} \)
Show Answer & Explanation

Answer: (a) \( \frac{7}{17} \)

Explanation:
1. Multiples of 3: 3, 6, 9, 12, 15 (5). Multiples of 7: 7, 14 (2). None common. Favourable: 7. P = 7/17.

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).

Question 2: A bag contains 5 green, 6 black and 7 white balls. A ball is drawn at random. The probability that it is white is:
(a) \( \frac{11}{18} \)
(b) \( \frac{4}{9} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{7}{18} \)
Show Answer & Explanation

Answer: (d) \( \frac{7}{18} \)

Explanation:
1. Total balls = 18, white = 7. P = 7/18.

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).

Question 3: A bag contains 5 green, 6 black and 7 white balls. The probability that the ball drawn is either green or black is:
(a) \( \frac{7}{18} \)
(b) \( \frac{2}{3} \)
(c) \( \frac{5}{9} \)
(d) \( \frac{11}{18} \)
Show Answer & Explanation

Answer: (d) \( \frac{11}{18} \)

Explanation:
1. Green or black = 5 + 6 = 11 of 18. P = 11/18.

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).

Question 4: A bag contains 5 green, 6 black and 7 white balls. The probability that the ball drawn is not black is:
(a) \( \frac{1}{3} \)
(b) \( \frac{13}{18} \)
(c) \( \frac{2}{3} \)
(d) \( \frac{11}{18} \)
Show Answer & Explanation

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

Explanation:
1. Not black = 5 + 7 = 12 of 18. P = 12/18 = 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).

Question 5: A bag contains 4 red and 8 blue marbles. A marble is drawn at random. The probability that it is red is:
(a) \( \frac{2}{3} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{5}{12} \)
(d) \( \frac{1}{4} \)
Show Answer & Explanation

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

Explanation:
1. 4 of the 12 marbles are red. P = 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).

Question 6: A bag contains 6 black, 7 red and 2 white balls. The probability that the ball drawn is red is:
(a) \( \frac{7}{15} \)
(b) \( \frac{8}{15} \)
(c) \( \frac{2}{5} \)
(d) \( \frac{7}{16} \)
Show Answer & Explanation

Answer: (a) \( \frac{7}{15} \)

Explanation:
1. 7 of the 15 balls are red. P = 7/15.

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).

Question 7: A bag contains 6 black, 7 red and 2 white balls. The probability that the ball drawn is black or white is:
(a) \( \frac{8}{15} \)
(b) \( \frac{7}{15} \)
(c) \( \frac{3}{5} \)
(d) \( \frac{1}{2} \)
Show Answer & Explanation

Answer: (a) \( \frac{8}{15} \)

Explanation:
1. Black or white = 6 + 2 = 8 of 15. P = 8/15.

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).

Question 8: A bag contains 6 black, 7 red and 2 white balls. The probability that the ball drawn is not black is:
(a) \( \frac{2}{5} \)
(b) \( \frac{2}{3} \)
(c) \( \frac{8}{15} \)
(d) \( \frac{3}{5} \)
Show Answer & Explanation

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

Explanation:
1. Not black = 7 + 2 = 9 of 15. P = 9/15 = 3/5.

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).

Question 9: A box contains 17 cards numbered 1, 2, ..., 17. A card is drawn at random. The probability that the number is odd is:
(a) \( \frac{8}{17} \)
(b) \( \frac{10}{17} \)
(c) \( \frac{9}{17} \)
(d) \( \frac{1}{2} \)
Show Answer & Explanation

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

Explanation:
1. The odd numbers 1, 3, ..., 17 are 9 in number. P = 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).

Question 10: A box contains 17 cards numbered 1 to 17. The probability that the number on the card drawn is prime is:
(a) \( \frac{10}{17} \)
(b) \( \frac{7}{17} \)
(c) \( \frac{8}{17} \)
(d) \( \frac{6}{17} \)
Show Answer & Explanation

Answer: (b) \( \frac{7}{17} \)

Explanation:
1. The primes are 2, 3, 5, 7, 11, 13, 17 (7). P = 7/17.

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).

Question 11: A box contains 17 cards numbered 1 to 17. The probability that the number on the card drawn is divisible by 3 is:
(a) \( \frac{12}{17} \)
(b) \( \frac{5}{17} \)
(c) \( \frac{6}{17} \)
(d) \( \frac{4}{17} \)
Show Answer & Explanation

Answer: (b) \( \frac{5}{17} \)

Explanation:
1. The numbers are 3, 6, 9, 12, 15 (5). P = 5/17.

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).

Question 12: A box contains 17 cards numbered 1 to 17. The probability that the number is not divisible by both 3 and 2 is:
(a) \( \frac{15}{17} \)
(b) \( \frac{2}{17} \)
(c) \( \frac{16}{17} \)
(d) \( \frac{14}{17} \)
Show Answer & Explanation

Answer: (a) \( \frac{15}{17} \)

Explanation:
1. Numbers divisible by both 3 and 2 (multiples of 6) are 6 and 12. Not divisible by both: 17 - 2 = 15. P = 15/17.

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).

Question 13: A box contains 11 cards numbered 1, 2, ..., 11. The probability that the number on the card drawn is prime is:
(a) \( \frac{6}{11} \)
(b) \( \frac{4}{11} \)
(c) \( \frac{5}{12} \)
(d) \( \frac{5}{11} \)
Show Answer & Explanation

Answer: (d) \( \frac{5}{11} \)

Explanation:
1. The primes are 2, 3, 5, 7, 11 (5). P = 5/11.

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).

Question 14: A box contains 11 cards numbered 1 to 11. The probability that the number is divisible by 3 is:
(a) \( \frac{8}{11} \)
(b) \( \frac{4}{11} \)
(c) \( \frac{3}{11} \)
(d) \( \frac{2}{11} \)
Show Answer & Explanation

Answer: (c) \( \frac{3}{11} \)

Explanation:
1. The numbers are 3, 6, 9 (3). P = 3/11.

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).

Question 15: A bag contains 4 blue balls and 3 red balls. A ball is drawn at random. The probability that it is not blue is:
(a) \( \frac{4}{7} \)
(b) \( \frac{2}{7} \)
(c) \( \frac{3}{7} \)
(d) \( \frac{3}{8} \)
Show Answer & Explanation

Answer: (c) \( \frac{3}{7} \)

Explanation:
1. Not blue means red: 3 of 7. P = 3/7.

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).

Question 16: A number is selected at random from the numbers 1 to 25. The probability that it is not a prime number is:
(a) \( \frac{9}{25} \)
(b) \( \frac{16}{25} \)
(c) \( \frac{17}{25} \)
(d) \( \frac{3}{5} \)
Show Answer & Explanation

Answer: (b) \( \frac{16}{25} \)

Explanation:
1. The primes up to 25 are 2, 3, 5, 7, 11, 13, 17, 19, 23 (9). Not prime: 25 - 9 = 16. P = 16/25.

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).

Question 17: A bag contains 5 red, 8 white, 4 green and 7 black balls. A ball is drawn at random. The probability that it is not green is:
(a) \( \frac{5}{6} \)
(b) \( \frac{1}{6} \)
(c) \( \frac{7}{8} \)
(d) \( \frac{19}{24} \)
Show Answer & Explanation

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

Explanation:
1. Total = 24, not green = 20. P = 20/24 = 5/6.

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).

Question 18: A piggy bank contains hundred 50 p coins, fifty Re 1 coins, twenty Rs 2 coins and ten Rs 5 coins. If one coin falls out at random, the probability that it will be a 50 p coin is:
(a) \( \frac{4}{9} \)
(b) \( \frac{101}{180} \)
(c) \( \frac{11}{20} \)
(d) \( \frac{5}{9} \)
Show Answer & Explanation

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

Explanation:
1. There are 100 + 50 + 20 + 10 = 180 coins and 100 are 50 p coins. P = 100/180 = 5/9.

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).

Question 19: A piggy bank contains hundred 50 p coins, fifty Re 1 coins, twenty Rs 2 coins and ten Rs 5 coins. The probability that the coin that falls out will not be a Rs 5 coin is:
(a) \( \frac{1}{18} \)
(b) \( \frac{19}{20} \)
(c) \( \frac{169}{180} \)
(d) \( \frac{17}{18} \)
Show Answer & Explanation

Answer: (d) \( \frac{17}{18} \)

Explanation:
1. 170 of the 180 coins are not Rs 5 coins. P = 170/180 = 17/18.

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).

Question 20: A lot of 20 bulbs contains 4 defective ones. One bulb is drawn at random. The probability that it is defective is:
(a) \( \frac{4}{5} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{1}{5} \)
(d) \( \frac{3}{20} \)
Show Answer & Explanation

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

Explanation:
1. P = 4/20 = 1/5.

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).

Question 21: A lot of 20 bulbs contains 4 defective ones. One bulb is drawn and found not defective and is not replaced. The probability that a second bulb drawn is not defective is:
(a) \( \frac{4}{19} \)
(b) \( \frac{15}{19} \)
(c) \( \frac{16}{19} \)
(d) \( \frac{14}{19} \)
Show Answer & Explanation

Answer: (b) \( \frac{15}{19} \)

Explanation:
1. Remaining bulbs: 19, of which 15 are not defective. P = 15/19.

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).

Question 22: An arrow rests at one of the numbers 1 to 8, all equally likely. The probability that it points at an odd number is:
(a) \( \frac{1}{2} \)
(b) \( \frac{5}{8} \)
(c) \( \frac{3}{8} \)
(d) \( \frac{4}{9} \)
Show Answer & Explanation

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

Explanation:
1. The odd numbers 1, 3, 5, 7 are 4 in number. P = 4/8 = 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).

Question 23: An arrow rests at one of the numbers 1 to 8, all equally likely. The probability that it points at a number greater than 2 is:
(a) \( \frac{3}{4} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{7}{8} \)
(d) \( \frac{5}{8} \)
Show Answer & Explanation

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

Explanation:
1. The numbers 3 to 8 are 6 in number. P = 6/8 = 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).

Question 24: A box contains 20 balls numbered 1 to 20. A ball is drawn at random. The probability that the number is divisible by 2 or 3 is:
(a) \( \frac{7}{20} \)
(b) \( \frac{7}{10} \)
(c) \( \frac{3}{5} \)
(d) \( \frac{13}{20} \)
Show Answer & Explanation

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

Explanation:
1. Divisible by 2: 10 numbers; by 3: 6 numbers; by both (multiples of 6): 3. So 10 + 6 - 3 = 13. 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).

Question 25: A box contains 20 balls numbered 1 to 20. The probability that the number on the ball is prime is:
(a) \( \frac{3}{5} \)
(b) \( \frac{9}{20} \)
(c) \( \frac{2}{5} \)
(d) \( \frac{7}{20} \)
Show Answer & Explanation

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

Explanation:
1. The primes are 2, 3, 5, 7, 11, 13, 17, 19 (8). 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).

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FAQs

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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 18 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 18, 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.

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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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