Question 1

A bag contains 5 red balls, 3 blue balls, and 2 green balls. If one ball is drawn at random from the bag, what is the probability that it is neither red nor green?

Accepted Answer

Step 1: Count total balls = 5 + 3 + 2 = 10 balls. Step 2: Identify favorable outcomes (neither red nor green) = blue balls = 3. Step 3: Apply probability formula: P(event) = Favorable outcomes / Total outcomes = 3/10. Therefore, the probability is 3/10 or 0.3.

Question 2

A bag contains 5 red balls, 3 blue balls, and 2 green balls. If one ball is drawn at random from the bag, what is the probability that it is neither red nor green?

Accepted Answer

Step 1: Total number of balls = 5 + 3 + 2 = 10. Step 2: Balls that are neither red nor green = blue balls = 3. Step 3: Probability = (Number of favourable outcomes) / (Total number of outcomes) = 3/10. Therefore, the answer is 3/10.

Question 3

A bag contains 5 red balls, 7 blue balls, and 8 green balls. Two balls are drawn simultaneously at random from the bag. What is the probability that both balls are of different colours?

Accepted Answer

Step 1: Total balls = 5 + 7 + 8 = 20. Total ways to draw 2 balls = C(20,2) = 190. Step 2: Ways to draw 2 balls of same colour = C(5,2) + C(7,2) + C(8,2) = 10 + 21 + 28 = 59. Step 3: Ways to draw 2 balls of different colours = 190 - 59 = 131. Step 4: Probability = 131/190. Therefore, answer is 131/190.

Question 4

A bag contains 4 red balls, 5 blue balls, and 3 green balls. Two balls are drawn at random without replacement. What is the probability that both balls are of the same colour?

Accepted Answer

Step 1: Total balls = 4 + 5 + 3 = 12. Total ways to choose 2 balls = C(12,2) = 66. Step 2: Favourable outcomes — both red: C(4,2) = 6; both blue: C(5,2) = 10; both green: C(3,2) = 3. Total favourable = 6 + 10 + 3 = 19. Step 3: Required probability = 19/66. Therefore, answer is 19/66.

Question 5

A bag contains 5 red balls, 7 blue balls, and 8 green balls. Two balls are drawn simultaneously at random from the bag. What is the probability that both balls are of different colours?

Accepted Answer

Step 1: Total balls = 5 + 7 + 8 = 20. Total ways to draw 2 balls = C(20,2) = 190. Step 2: Ways to draw both balls of same colour = C(5,2) + C(7,2) + C(8,2) = 10 + 21 + 28 = 59. Step 3: Ways to draw balls of different colours = 190 - 59 = 131. Step 4: Probability = 131/190. Therefore, answer is 131/190.

Question 6

A coin is tossed 3 times. What is the probability of getting exactly 2 heads?

Accepted Answer

Step 1: Total possible outcomes when tossing a coin 3 times = 2³ = 8. Step 2: Favorable outcomes (exactly 2 heads): HHT, HTH, THH = 3 outcomes. Step 3: Probability = 3/8. Therefore, the answer is 3/8.

Question 7

A bag contains 5 red balls, 3 blue balls, and 2 green balls. If one ball is drawn at random, what is the probability that it is either red or green?

Accepted Answer

Step 1: Total number of balls = 5 + 3 + 2 = 10. Step 2: Number of red or green balls = 5 + 2 = 7. Step 3: Probability = Favourable outcomes / Total outcomes = 7/10 = 0.7. Therefore, the answer is 7/10.

Question 8

A coin is flipped 3 times. What is the probability of getting exactly 2 heads?

Accepted Answer

Step 1: Total possible outcomes when flipping a coin 3 times = 2³ = 8. Step 2: Favorable outcomes (exactly 2 heads): HHT, HTH, THH = 3 outcomes. Step 3: Probability = 3/8. Therefore, the answer is 3/8.

Question 9

A bag contains 5 red balls, 3 blue balls, and 2 green balls. If one ball is drawn at random, what is the probability that it is either red or green?

Accepted Answer

Step 1: Total number of balls = 5 + 3 + 2 = 10. Step 2: Number of red or green balls = 5 + 2 = 7. Step 3: Probability = Favorable outcomes / Total outcomes = 7/10 = 0.7. Therefore, the answer is 7/10.

Question 10

A student has a 60% chance of passing Mathematics and a 75% chance of passing English. Assuming independence, what is the probability that the student passes at least one subject?

Accepted Answer

Step 1: P(pass Math) = 0.60, P(pass English) = 0.75. Step 2: P(fail Math) = 0.40, P(fail English) = 0.25. Step 3: P(fail both) = 0.40 × 0.25 = 0.10. Step 4: P(pass at least one) = 1 - P(fail both) = 1 - 0.10 = 0.90 = 9/10.

Question 11

In a lottery, the probability of winning a prize is 0.12. If 500 tickets are sold, how many tickets are expected to win a prize?

Accepted Answer

Step 1: Probability of winning = 0.12. Step 2: Number of tickets = 500. Step 3: Expected number of winning tickets = Probability × Total tickets = 0.12 × 500 = 60.

Question 12

A box contains 6 defective and 14 non-defective items. If 3 items are drawn at random without replacement, what is the probability that all 3 are non-defective?

Accepted Answer

Step 1: Total items = 20. Step 2: Probability using sequential draws without replacement: P = (14/20) × (13/19) × (12/18). Step 3: = (14×13×12)/(20×19×18) = 2184/6840. Step 4: Simplify by GCD(2184, 6840) = 24: 2184/24 = 91, 6840/24 = 285. So P = 91/285.

Question 13

A bag contains 5 red balls, 7 blue balls, and 8 green balls. If two balls are drawn simultaneously at random, what is the probability that both balls are of the same colour?

Accepted Answer

Step 1: Total balls = 5 + 7 + 8 = 20. Total ways to draw 2 balls = C(20,2) = 190. Step 2: Ways to draw 2 red balls = C(5,2) = 10. Ways to draw 2 blue balls = C(7,2) = 21. Ways to draw 2 green balls = C(8,2) = 28. Step 3: Favourable outcomes = 10 + 21 + 28 = 59. Step 4: Probability = 59/190.

Question 14

A bag contains 5 red balls, 7 blue balls, and 8 green balls. If two balls are drawn simultaneously at random, what is the probability that both balls are of the same colour?

Accepted Answer

Step 1: Total balls = 5 + 7 + 8 = 20. Total ways to draw 2 balls = C(20,2) = 190. Step 2: Ways to draw 2 red balls = C(5,2) = 10. Ways to draw 2 blue balls = C(7,2) = 21. Ways to draw 2 green balls = C(8,2) = 28. Step 3: Favourable outcomes = 10 + 21 + 28 = 59. Step 4: Probability = 59/190.

Question 15

In a lottery, the probability of winning a prize is 0.15. If 200 tickets are sold and a person buys 8 tickets, what is the probability that this person wins exactly 2 prizes?

Accepted Answer

Step 1: This is a binomial probability problem with n = 8, p = 0.15, q = 0.85, and r = 2. Step 2: P(X = 2) = C(8,2) × (0.15)² × (0.85)⁶. Step 3: C(8,2) = 28. Step 4: (0.15)² = 0.0225. Step 5: (0.85)⁶ ≈ 0.3771. Step 6: P(X = 2) = 28 × 0.0225 × 0.3771 ≈ 0.2376.

Question 16

A bag contains 5 red balls, 7 blue balls, and 8 green balls. Two balls are drawn simultaneously without replacement. What is the probability that both balls are of different colours?

Accepted Answer

Step 1: Total balls = 5 + 7 + 8 = 20. Total ways to draw 2 balls = C(20,2) = 190. Step 2: Ways to draw both same colour = C(5,2) + C(7,2) + C(8,2) = 10 + 21 + 28 = 59. Step 3: Ways to draw different colours = 190 - 59 = 131. Step 4: Probability = 131/190. Therefore, answer is 131/190.

Question 17

A bag contains 5 red balls, 7 blue balls, and 8 green balls. Two balls are drawn simultaneously without replacement. What is the probability that both balls are of different colours?

Accepted Answer

Step 1: Total balls = 5 + 7 + 8 = 20. Total ways to draw 2 balls = C(20,2) = 190. Step 2: Ways to draw both same colour = C(5,2) + C(7,2) + C(8,2) = 10 + 21 + 28 = 59. Step 3: Ways to draw different colours = 190 - 59 = 131. Step 4: Probability = 131/190.

Question 18

Three fair dice are rolled simultaneously. What is the probability that the sum of the numbers shown is 10, given that at least one die shows a 4?

Accepted Answer

Step 1: Find total outcomes where at least one die shows 4. Using complement: Total outcomes = 6³ = 216. Outcomes with NO 4 = 5³ = 125. Outcomes with at least one 4 = 216 - 125 = 91. Step 2: Find outcomes where sum = 10 AND at least one die shows 4. Possible combinations summing to 10: (2,4,4), (4,2,4), (4,4,2), (3,3,4), (3,4,3), (4,3,3), (1,5,4), (1,4,5), (5,1,4), (5,4,1), (4,1,5), (4,5,1), (2,3,5), (2,5,3), (3,2,5), (3,5,2), (5,2,3), (5,3,2). Filter those with at least one 4: (2,4,4), (4,2,4), (4,4,2), (3,3,4), (3,4,3), (4,3,3), (1,5,4), (1,4,5), (5,1,4), (5,4,1), (4,1,5), (4,5,1) = 12 outcomes. Step 3: Conditional probability = 12/91.

RRB Ministerial & Isolated Probability

RRB Ministerial & Isolated Probability — Past Exam Questions

Probability— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Probability — Practice Questions

60-Second Revision — Probability