CBSE Class 10 Mathematics Probability MCQs Set 17

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

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Multiple Choice Questions

Question 1: A card is drawn from a pack of 52 cards. The probability of getting an ace is:
(a) \( \frac{1}{13} \)
(b) \( \frac{12}{13} \)
(c) \( \frac{5}{52} \)
(d) \( \frac{3}{52} \)
Show Answer & Explanation

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

Explanation:
1. There are 4 aces among 52 cards, so P = 4/52 = 1/13.

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).

Question 2: A card is drawn from a pack of 52 cards. The probability that it is either a king or a queen is:
(a) \( \frac{11}{13} \)
(b) \( \frac{9}{52} \)
(c) \( \frac{7}{52} \)
(d) \( \frac{2}{13} \)
Show Answer & Explanation

Answer: (d) \( \frac{2}{13} \)

Explanation:
1. There are 4 kings and 4 queens, 8 favourable cards. P = 8/52 = 2/13.

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).

Question 3: One card is drawn from a pack of 52 cards. The probability that it is red or a king is:
(a) \( \frac{6}{13} \)
(b) \( \frac{29}{52} \)
(c) \( \frac{7}{13} \)
(d) \( \frac{27}{52} \)
Show Answer & Explanation

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

Explanation:
1. Red cards: 26. Kings: 4, of which 2 are red. Red or king = 26 + 4 - 2 = 28. P = 28/52 = 7/13.

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).

Question 4: The king, queen and jack of clubs are removed from a pack of 52 cards and the rest is shuffled. One card is drawn from the remaining 49. The probability that it is a heart is:
(a) \( \frac{36}{49} \)
(b) \( \frac{13}{49} \)
(c) \( \frac{2}{7} \)
(d) \( \frac{12}{49} \)
Show Answer & Explanation

Answer: (b) \( \frac{13}{49} \)

Explanation:
1. All 13 hearts remain among 49 cards. P = 13/49.

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).

Question 5: The king, queen and jack of clubs are removed from a pack of 52 cards and one card is drawn from the remaining 49. The probability that it is a king 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. Three kings remain (the king of clubs was removed). 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).

Question 6: The king, queen and jack of clubs are removed from a pack of 52 cards and one card is drawn from the remaining 49. The probability that it is a club is:
(a) \( \frac{10}{49} \)
(b) \( \frac{39}{49} \)
(c) \( \frac{11}{49} \)
(d) \( \frac{9}{49} \)
Show Answer & Explanation

Answer: (a) \( \frac{10}{49} \)

Explanation:
1. Clubs remaining: 13 - 3 = 10. P = 10/49.

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).

Question 7: The king, queen and jack of clubs are removed from a pack of 52 cards and one card is drawn from the remaining 49. The probability that it is the 10 of hearts is:
(a) \( \frac{48}{49} \)
(b) \( \frac{2}{49} \)
(c) \( 0 \)
(d) \( \frac{1}{49} \)
Show Answer & Explanation

Answer: (d) \( \frac{1}{49} \)

Explanation:
1. There is exactly one 10 of hearts among 49 cards. P = 1/49.

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).

Question 8: A card is drawn from a pack of 52 cards. The probability that it is a black king is:
(a) \( \frac{25}{26} \)
(b) \( \frac{3}{52} \)
(c) \( \frac{1}{26} \)
(d) \( \frac{1}{52} \)
Show Answer & Explanation

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

Explanation:
1. The black kings are the king of spades and the king of clubs, 2 cards. 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).

Question 9: A card is drawn from a pack of 52 cards. The probability that it is either a black card or a king is:
(a) \( \frac{6}{13} \)
(b) \( \frac{29}{52} \)
(c) \( \frac{7}{13} \)
(d) \( \frac{27}{52} \)
Show Answer & Explanation

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

Explanation:
1. Black cards: 26. Kings: 4, of which 2 are black. Black or king = 26 + 4 - 2 = 28. P = 28/52 = 7/13.

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).

Question 10: A card is drawn from a pack of 52 cards. The probability that it is a jack, queen or king is:
(a) \( \frac{10}{13} \)
(b) \( \frac{3}{13} \)
(c) \( \frac{1}{4} \)
(d) \( \frac{11}{52} \)
Show Answer & Explanation

Answer: (b) \( \frac{3}{13} \)

Explanation:
1. There are 3 x 4 = 12 such cards. P = 12/52 = 3/13.

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).

Question 11: A card is drawn from a pack of 52 cards. The probability that it is neither a heart nor a king is:
(a) \( \frac{9}{13} \)
(b) \( \frac{4}{13} \)
(c) \( \frac{37}{52} \)
(d) \( \frac{35}{52} \)
Show Answer & Explanation

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

Explanation:
1. Hearts or kings: 13 + 4 - 1 = 16. Neither: 52 - 16 = 36. P = 36/52 = 9/13.

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).

Question 12: A card is drawn from a pack of 52 cards. The probability that it is a spade or an ace is:
(a) \( \frac{9}{13} \)
(b) \( \frac{17}{52} \)
(c) \( \frac{15}{52} \)
(d) \( \frac{4}{13} \)
Show Answer & Explanation

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

Explanation:
1. Spades: 13. Aces: 4, of which 1 is a spade. Spade or ace = 13 + 4 - 1 = 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).

Question 13: A card is drawn from a pack of 52 cards. The probability that it is neither an ace nor a king is:
(a) \( \frac{2}{13} \)
(b) \( \frac{45}{52} \)
(c) \( \frac{43}{52} \)
(d) \( \frac{11}{13} \)
Show Answer & Explanation

Answer: (d) \( \frac{11}{13} \)

Explanation:
1. Aces and kings: 8 cards. Neither: 44. P = 44/52 = 11/13.

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).

Question 14: A card is drawn from a pack of 52 cards. The probability that it is a red queen is:
(a) \( \frac{25}{26} \)
(b) \( \frac{3}{52} \)
(c) \( \frac{1}{26} \)
(d) \( \frac{1}{52} \)
Show Answer & Explanation

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

Explanation:
1. The red queens are the queen of hearts and the queen of diamonds. 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).

Question 15: A card is drawn from a pack of 52 cards. The probability that it is a black face card is:
(a) \( \frac{23}{26} \)
(b) \( \frac{3}{26} \)
(c) \( \frac{7}{52} \)
(d) \( \frac{5}{52} \)
Show Answer & Explanation

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

Explanation:
1. Black face cards: 3 in spades and 3 in clubs = 6. P = 6/52 = 3/26.

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).

Question 16: A card is drawn from a pack of 52 cards. The probability that it is the 2 of spades is:
(a) \( \frac{1}{52} \)
(b) \( \frac{51}{52} \)
(c) \( \frac{1}{26} \)
(d) \( 0 \)
Show Answer & Explanation

Answer: (a) \( \frac{1}{52} \)

Explanation:
1. There is exactly one such card. P = 1/52.

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 card is drawn from a pack of 52 cards. The probability that it is a 10 of a black suit 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. The 10 of spades and the 10 of clubs: 2 cards. 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).

Question 18: A card is drawn at random from a well-shuffled pack of 52 cards. The probability that it is neither a red card nor a queen is:
(a) \( \frac{7}{13} \)
(b) \( \frac{25}{52} \)
(c) \( \frac{23}{52} \)
(d) \( \frac{6}{13} \)
Show Answer & Explanation

Answer: (d) \( \frac{6}{13} \)

Explanation:
1. Red cards: 26 plus the 2 black queens = 28 cards that are red or a queen. Neither: 52 - 28 = 24. P = 24/52 = 6/13.

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).

Question 19: A coin is tossed two times. The probability of getting a head at least once 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. Outcomes: HH, HT, TH, TT. Only TT has no head, so 3 of 4. 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).

Question 20: Two coins are tossed simultaneously. The probability of getting two tails is:
(a) \( \frac{3}{4} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{1}{2} \)
(d) \( 0 \)
Show Answer & Explanation

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

Explanation:
1. Only TT is favourable out of 4 outcomes. P = 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).

Question 21: Two coins are tossed simultaneously. The probability of getting no tail is:
(a) \( \frac{3}{4} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{1}{2} \)
(d) \( 0 \)
Show Answer & Explanation

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

Explanation:
1. Only HH is favourable out of 4 outcomes. P = 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).

Question 22: Three coins are tossed simultaneously. The probability of getting exactly 2 tails is:
(a) \( \frac{3}{8} \)
(b) \( \frac{5}{8} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{1}{4} \)
Show Answer & Explanation

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

Explanation:
1. The outcomes with exactly two tails are TTH, THT, HTT: 3 of 8. P = 3/8.

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: Three coins are tossed simultaneously. The probability of getting 3 tails is:
(a) \( \frac{7}{8} \)
(b) \( \frac{1}{4} \)
(c) \( 0 \)
(d) \( \frac{1}{8} \)
Show Answer & Explanation

Answer: (d) \( \frac{1}{8} \)

Explanation:
1. Only TTT is favourable out of 8 outcomes. P = 1/8.

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: Three coins are tossed simultaneously. The probability of getting two heads and one tail is:
(a) \( \frac{5}{8} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{3}{8} \)
(d) \( \frac{1}{4} \)
Show Answer & Explanation

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

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

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: Three coins are tossed simultaneously. The probability of getting at least one head is:
(a) \( \frac{1}{8} \)
(b) \( 1 \)
(c) \( \frac{7}{8} \)
(d) \( \frac{3}{4} \)
Show Answer & Explanation

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

Explanation:
1. Only TTT has no head. P = 1 - 1/8 = 7/8.

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 26: A number is chosen at random from the first 100 natural numbers. The probability that it is a multiple of 5 is:
(a) \( \frac{4}{5} \)
(b) \( \frac{1}{5} \)
(c) \( \frac{21}{100} \)
(d) \( \frac{19}{100} \)
Show Answer & Explanation

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

Explanation:
1. Multiples of 5 up to 100: 20. P = 20/100 = 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).

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