CBSE Class 10 Mathematics Probability MCQs Set 14

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

Access the complete set of multiple-choice questions for Chapter 14 Probability below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official CBSE textbooks.

Multiple Choice Questions

Question 1: A letter is chosen at random from the English alphabet. Find the probability that the letter chosen precedes g.
(a) \( \frac{7}{26} \)
(b) \( \frac{3}{13} \)
(c) \( \frac{1}{26} \)
(d) \( \frac{10}{13} \)
Show Answer & Explanation

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

Explanation:
1. Letters before g: a, b, c, d, e, f, which are 6.
2. Total letters = 26.
3. P = 6/26 = 3/13.

Teacher's Note:
1. Letters before g are a, b, c, d, e, f.
2. Count them carefully; g itself is not included.

Question 2: An integer is chosen between 0 and 100. What is the probability that it is divisible by 7?
(a) \( \frac{14}{99} \)
(b) \( \frac{3}{20} \)
(c) \( \frac{7}{50} \)
(d) \( \frac{1}{7} \)
Show Answer & Explanation

Answer: (a) \( \frac{14}{99} \)

Explanation:
1. Integers between 0 and 100: 1 to 99, which are 99.
2. Multiples of 7: 7, 14, ..., 98, which are 14.
3. P = 14/99.

Teacher's Note:
1. Integers between 0 and 100 means 1 to 99 only.
2. Count multiples of 7 as 7 x 1 up to 7 x 14.

Question 3: A bag contains 7 red, 5 white and 3 black balls. A ball is drawn at random. Find the probability that it is neither white nor black.
(a) \( \frac{8}{15} \)
(b) \( \frac{7}{16} \)
(c) \( \frac{2}{5} \)
(d) \( \frac{7}{15} \)
Show Answer & Explanation

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

Explanation:
1. Total balls = 15.
2. Neither white nor black means red, so 7 balls.
3. P = 7/15.

Teacher's Note:
1. Neither white nor black leaves only the red balls.
2. Add all balls first to get the total.

Question 4: Bag X has four cards numbered 1, 3, 5, 7 and bag Y has three cards numbered 2, 4, 6. One card is drawn at random from each bag. Find the probability that the sum of the two numbers is 7.
(a) \( \frac{3}{4} \)
(b) \( \frac{3}{13} \)
(c) \( \frac{1}{4} \)
(d) \( \frac{1}{3} \)
Show Answer & Explanation

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

Explanation:
1. Total outcomes = 4 x 3 = 12.
2. Sum 7: (1,6), (3,4), (5,2), which are 3.
3. P = 3/12 = 1/4.

Teacher's Note:
1. Two bags give 4 x 3 = 12 equally likely pairs.
2. List pairs with sum 7 systematically.

Question 5: Bag X has four cards numbered 1, 3, 5, 7 and bag Y has three cards numbered 2, 4, 6. One card is drawn at random from each bag. Find the probability that the sum of the two numbers is more than 7.
(a) \( \frac{6}{13} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{7}{12} \)
(d) \( \frac{5}{12} \)
Show Answer & Explanation

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

Explanation:
1. Total outcomes = 4 x 3 = 12.
2. Sums more than 7 (as card pairs): (3,6), (5,4), (5,6), (7,2), (7,4), (7,6), which are 6.
3. P = 6/12 = 1/2.

Teacher's Note:
1. Write all 12 pairs and their sums before counting.
2. Sum more than 7 means 9, 11 or 13 here.

Question 6: The king, queen and jack of diamonds are removed from a pack of 52 cards and the pack is well shuffled. A card is drawn from the remaining cards. Find the probability that it is a diamond.
(a) \( \frac{10}{49} \)
(b) \( \frac{39}{49} \)
(c) \( \frac{1}{5} \)
(d) \( \frac{11}{49} \)
Show Answer & Explanation

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

Explanation:
1. Cards left = 52 - 3 = 49.
2. Diamonds left = 13 - 3 = 10.
3. P = 10/49.

Teacher's Note:
1. Removing cards changes both the total and the diamonds count.
2. Diamonds left = 13 minus 3.

Question 7: The king, queen and jack of diamonds are removed from a pack of 52 cards and the pack is well shuffled. A card is drawn from the remaining cards. Find the probability that it is a jack.
(a) \( \frac{46}{49} \)
(b) \( \frac{3}{50} \)
(c) \( \frac{4}{49} \)
(d) \( \frac{3}{49} \)
Show Answer & Explanation

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

Explanation:
1. Cards left = 49.
2. Jacks left = 4 - 1 = 3.
3. P = 3/49.

Teacher's Note:
1. Only the jack of diamonds was removed from the four jacks.
2. Total cards left = 49.

Question 8: A bag contains tickets numbered 11, 12, 13, ..., 30. A ticket is taken out at random. Find the probability that the number on it is a multiple of 7.
(a) \( \frac{17}{20} \)
(b) \( \frac{1}{7} \)
(c) \( \frac{3}{20} \)
(d) \( \frac{1}{5} \)
Show Answer & Explanation

Answer: (c) \( \frac{3}{20} \)

Explanation:
1. Tickets: 11 to 30, so 20 tickets.
2. Multiples of 7: 14 and 21 and 28, which are 3.
3. P = 3/20.

Teacher's Note:
1. Count the multiples of 7 inside 11 to 30 only.
2. Total tickets = 30 - 11 + 1 = 20.

Question 9: A bag contains tickets numbered 11, 12, 13, ..., 30. A ticket is taken out at random. Find the probability that the number on it is greater than 15 and a multiple of 5.
(a) \( \frac{17}{20} \)
(b) \( \frac{3}{20} \)
(c) \( \frac{1}{7} \)
(d) \( \frac{1}{5} \)
Show Answer & Explanation

Answer: (b) \( \frac{3}{20} \)

Explanation:
1. Tickets: 20 in all.
2. Multiples of 5 greater than 15: 20, 25, 30, which are 3.
3. P = 3/20.

Teacher's Note:
1. The conditions greater than 15 and multiple of 5 must both hold.
2. Check 15 is excluded since it is not greater than 15.

Question 10: Two dice are thrown simultaneously. What is the probability that 5 will come up on both dice?
(a) \( \frac{1}{36} \)
(b) \( \frac{35}{36} \)
(c) \( \frac{1}{37} \)
(d) \( \frac{1}{18} \)
Show Answer & Explanation

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

Explanation:
1. Only the outcome (5, 5) is favourable.
2. P = 1/36.

Teacher's Note:
1. Only one of the 36 outcomes is (5, 5).
2. On two dice, (a, b) and (b, a) are different outcomes.

Question 11: Red kings, queens and jacks are removed from a deck of 52 playing cards and the rest are well shuffled. A card is drawn from the remaining cards. Find the probability of getting a king.
(a) \( \frac{22}{23} \)
(b) \( \frac{2}{47} \)
(c) \( \frac{3}{46} \)
(d) \( \frac{1}{23} \)
Show Answer & Explanation

Answer: (d) \( \frac{1}{23} \)

Explanation:
1. Cards left = 52 - 6 = 46.
2. Kings left = 2 (black).
3. P = 2/46 = 1/23.

Teacher's Note:
1. Only red kings are removed, so the two black kings remain.
2. Total cards left = 52 - 6 = 46.

Question 12: Red kings, queens and jacks are removed from a deck of 52 playing cards and the rest are well shuffled. A card is drawn from the remaining cards. Find the probability of getting a red card.
(a) \( \frac{13}{23} \)
(b) \( \frac{20}{47} \)
(c) \( \frac{10}{23} \)
(d) \( \frac{21}{46} \)
Show Answer & Explanation

Answer: (c) \( \frac{10}{23} \)

Explanation:
1. Cards left = 46.
2. Red cards left = 26 - 6 = 20.
3. P = 20/46 = 10/23.

Teacher's Note:
1. Red cards left = 26 minus 6 removed.
2. The total is now 46, not 52.

Question 13: Red kings, queens and jacks are removed from a deck of 52 playing cards and the rest are well shuffled. A card is drawn from the remaining cards. Find the probability of getting a spade.
(a) \( \frac{33}{46} \)
(b) \( \frac{13}{46} \)
(c) \( \frac{13}{47} \)
(d) \( \frac{7}{23} \)
Show Answer & Explanation

Answer: (b) \( \frac{13}{46} \)

Explanation:
1. Cards left = 46.
2. Spades are all black, so all 13 remain.
3. P = 13/46.

Teacher's Note:
1. Spades are black, so none of them is removed.
2. Keep the total as 46.

Question 14: A pair of dice is thrown once. Find the probability of getting an odd number on each die.
(a) \( \frac{1}{4} \)
(b) \( \frac{3}{4} \)
(c) \( \frac{9}{37} \)
(d) \( \frac{5}{18} \)
Show Answer & Explanation

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

Explanation:
1. Odd numbers on a die: 1, 3, 5, so 3 choices each.
2. Favourable = 3 x 3 = 9 out of 36.
3. P = 9/36 = 1/4.

Teacher's Note:
1. Each die has 3 odd faces: 1, 3, 5.
2. Multiply the choices: 3 x 3 = 9 favourable outcomes.

Question 15: A lot consists of 48 mobile phones of which 42 are good, 3 have only minor defects and 3 have major defects. Varnika buys a phone if it is good. One phone is selected at random. What is the probability that it is acceptable to Varnika?
(a) \( \frac{1}{8} \)
(b) \( \frac{6}{7} \)
(c) \( \frac{43}{48} \)
(d) \( \frac{7}{8} \)
Show Answer & Explanation

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

Explanation:
1. Acceptable to Varnika = good phones = 42.
2. P = 42/48 = 7/8.

Teacher's Note:
1. Acceptable to Varnika means only the good phones.
2. Total phones = 48.

Question 16: A lot consists of 48 mobile phones of which 42 are good, 3 have only minor defects and 3 have major defects. A trader buys a phone only if it has no major defect. One phone is selected at random. What is the probability that it is acceptable to the trader?
(a) \( \frac{1}{16} \)
(b) \( \frac{45}{49} \)
(c) \( \frac{15}{16} \)
(d) \( \frac{23}{24} \)
Show Answer & Explanation

Answer: (c) \( \frac{15}{16} \)

Explanation:
1. No major defect = 42 + 3 = 45.
2. P = 45/48 = 15/16.

Teacher's Note:
1. The trader accepts good phones and those with only minor defects.
2. Subtract only the 3 with major defects from 48.

Question 17: Find the probability of getting 53 Sundays in a leap year.
(a) \( \frac{5}{7} \)
(b) \( \frac{2}{7} \)
(c) \( \frac{1}{4} \)
(d) \( \frac{3}{7} \)
Show Answer & Explanation

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

Explanation:
1. A leap year has 366 days = 52 weeks + 2 days.
2. The two extra days can be (Sun,Mon), (Mon,Tue), (Tue,Wed), (Wed,Thu), (Thu,Fri), (Fri,Sat), (Sat,Sun).
3. Sunday is in 2 of these 7 pairs, so P = 2/7.

Teacher's Note:
1. A leap year has 52 full weeks and 2 extra days.
2. The extra two days are consecutive, giving 7 equally likely pairs.

Question 18: From a well-shuffled pack of playing cards, the black jacks, black kings and black aces are removed. A card is then drawn from the pack. Find the probability that it is a red card.
(a) \( \frac{13}{23} \)
(b) \( \frac{10}{23} \)
(c) \( \frac{26}{47} \)
(d) \( \frac{27}{46} \)
Show Answer & Explanation

Answer: (a) \( \frac{13}{23} \)

Explanation:
1. Removed cards = 2 + 2 + 2 = 6, so 46 remain.
2. All 26 red cards remain.
3. P = 26/46 = 13/23.

Teacher's Note:
1. The 6 removed cards are all black, so all 26 red cards remain.
2. Total cards left = 46.

Question 19: From a well-shuffled pack of playing cards, the black jacks, black kings and black aces are removed. A card is then drawn from the pack. Find the probability that it is not a diamond card.
(a) \( \frac{13}{46} \)
(b) \( \frac{33}{47} \)
(c) \( \frac{17}{23} \)
(d) \( \frac{33}{46} \)
Show Answer & Explanation

Answer: (d) \( \frac{33}{46} \)

Explanation:
1. Cards left = 46.
2. Diamonds = 13 (none removed), so not diamond = 46 - 13 = 33.
3. P = 33/46.

Teacher's Note:
1. No diamonds are removed, so 13 diamonds remain.
2. Not a diamond = 46 minus 13.

Question 20: Five cards, the ten, jack, queen, king and ace of diamonds, are well shuffled with their faces downwards. One card is picked up and put aside. A second card is then picked up. If the first card was a queen, what is the probability that the second card is an ace?
(a) \( \frac{3}{4} \)
(b) \( \frac{1}{5} \)
(c) \( \frac{1}{4} \)
(d) \( \frac{1}{2} \)
Show Answer & Explanation

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

Explanation:
1. After removing the queen, 4 cards remain.
2. The ace is one of them.
3. P = 1/4.

Teacher's Note:
1. After the queen is put aside, 4 cards remain.
2. The ace is still among them, so it is 1 favourable card out of 4.

Question 21: Two customers, Shyam and Ekta, visit a shop in the same week (Tuesday to Saturday). Each is equally likely to visit on any of the 5 days. What is the probability that both visit the shop on consecutive days?
(a) \( \frac{17}{25} \)
(b) \( \frac{8}{25} \)
(c) \( \frac{4}{13} \)
(d) \( \frac{9}{25} \)
Show Answer & Explanation

Answer: (b) \( \frac{8}{25} \)

Explanation:
1. Total outcomes = 5 x 5 = 25.
2. Consecutive-day pairs (in either order): (Tue,Wed), (Wed,Thu), (Thu,Fri), (Fri,Sat) and the 4 reversed ones, which are 8.
3. P = 8/25.

Teacher's Note:
1. There are 5 x 5 = 25 equally likely pairs of days.
2. Count consecutive-day pairs in both orders.

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Chapter 14 Probability Objective Questions & Solutions for Class 10 Mathematics

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FAQs

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