Question 1

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 2

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 3

In a class of 40 students, 15 play cricket, 20 play football, and 8 play both. If a student is selected at random, what is the probability that the student plays at least one sport?

Accepted Answer

Step 1: Using inclusion-exclusion: Students playing at least one sport = 15 + 20 - 8 = 27. Step 2: Total students = 40. Step 3: Probability = 27/40.

Question 4

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 5

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 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 6

In a lottery, 6 tickets are drawn from 100 tickets numbered 1 to 100. What is the probability that the sum of the numbers on the drawn tickets is even?

Accepted Answer

Step 1: For sum to be even, we need either all 6 even numbers, or exactly 4 even and 2 odd, or exactly 2 even and 4 odd, or all 6 odd. Step 2: There are 50 even and 50 odd numbers. Step 3: Total ways = C(100,6). Step 4: Even sum cases = C(50,6) + C(50,4)×C(50,2) + C(50,2)×C(50,4) + C(50,6). Step 5: This simplifies to C(50,6) + C(50,4)×C(50,2) + C(50,2)×C(50,4) + C(50,6) = 2×C(50,6) + 2×C(50,4)×C(50,2). Step 6: By symmetry, P(even sum) = 1/2. Therefore, answer is 1/2.

SSC CHSL Probability

Probability— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Probability — Practice Questions

60-Second Revision — Probability