Finite sales positions problem

Question

Assume that 12 people, including the husband and wife pair, apply for 5 sales positions. People are hired at random.

1) What is the probability that both the husband and wife are hired?

2) What is the probability that one is hired and one is not?

Answer 1

First, we need to know how many ways 5 of these 12 can be selected for the sales positions. Since order doesn't matter, we'll use combinations: ₁₂C₅ = ^12!/_5!•7! = 792. Then remember that probability is the number of ways you want to happen divided by this total.

1) How many ways can the husband and wife be selected? Well now we'll just select the husband and wife outright and then select 3 of the remaining 10 people. ₁₀C₃ = ^10!/_3!•7! = 120. This means that the probability that both the wife and husband are hired is ¹²⁰/₇₉₂ ≈ 15.152%.

2) How many ways can one of the two get hired and the other not. In this case, we'd first choose 1 of the 2 (₂C₁ = 2) and multiply this by the number of ways we can select 4 of the remaining 10 candidates.

2*₁₀C₄ = ^10!/_4!•6! = 210. This means that the probability that only one of this couple is selected is ²¹⁰/₇₉₂ ≈ 26.515%.