Question

A town recently dismissed 10 employees in order to meet their new budget reductions. The town had 7 employees over 50 years of age and 18 under 50. If the dismissed employees were selected at random, what is the probability that exactly 5 employees were over 50

Answer 1

0.055 = 5.5% probability that exactly 5 employees were over 50.

Explanation:

The employees are dismissed from the sample without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

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)!)`

In this question:

Total of 7 + 18 = 25 employees, which means that
`N = 25`

7 over 50, which means that
`k = 7`

10 were dismissed, which means that
`n = 10`

What is the probability that exactly 5 employees were over 50?

This is P(X = 5). So

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

`P(X = 5) = h(5,25,10,7) = (C_(7,5)*C_(18,5))/(C_(25,10)) = 0.055`

0.055 = 5.5% probability that exactly 5 employees were over 50.