IMSA Probability and Combinatorics 2

Question 1

How many ways can you choose 3 students from a class of 10 (order doesn't matter)?

Accepted Answer

120

Answer

C(10,3) = 10!/(3! × 7!) = (10 × 9 × 8)/(3 × 2 × 1) = 120.

Question 2

Two dice are rolled. What is the probability that both show a 6?

Accepted Answer

1/36

Answer

P(6) × P(6) = (1/6) × (1/6) = 1/36.

Question 3

A spinner has 8 equal sections numbered 1−8. What is the probability of landing on an odd number?

Accepted Answer

1/2

Answer

The odd numbers are 1, 3, 5, 7 (four values), so P = 4/8 = 1/2.

Question 4

How many different 3-letter arrangements can be made from the letters A, B, C, D (no repetition)?

Accepted Answer

24

Answer

This is a permutation: P(4,3) = 4 × 3 × 2 = 24.

Question 5

Events A and B are mutually exclusive. P(A) = 0.3 and P(B) = 0.4. What is P(A or B)?

Accepted Answer

0.7

Answer

For mutually exclusive events, P(A or B) = P(A) + P(B) = 0.3 + 0.4 = 0.7.

IMSA - Intermediate Math Skill Assesment Practice Test