Question

Assume that 5 people, including the husband and wife pair, apply for 4 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

a) 0.2 = 20% probability that both the husband and wife are hired

b) 0.4 = 40% probability that one is hired and one is not

Explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the people are chosen is not important, which means that we use the combinations formula to solve this question.

Combinations formula:

`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

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

Desired outcomes:

Both selected, only one possible outcome, so
`D = 1`

Total outcomes:

4 from a set of 5. So

`T = C_(5,4) = (5!)/(4!1!) = 5`

Probability:

`p = (D)/(T) = (1)/(5) = 0.2`

0.2 = 20% probability that both the husband and wife are hired

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

Desired outcomes:

One selected, one not, so two possible outcomes(Wife yes, husband not, or husband yes and wife not). So
`D = 2`

Total outcomes:

4 from a set of 5. So

`T = C_(5,4) = (5!)/(4!1!) = 5`

Probability:

`p = (D)/(T) = (2)/(5) = 0.4`

0.4 = 40% probability that one is hired and one is not