Question

A town recently dismissed 88 employees in order to meet their new budget reductions. The town had 77 employees over 5050 years of age and 1717 under 5050. If the dismissed employees were selected at random, what is the probability that at least 66 employees were over 5050? Express your answer as a fraction or a decimal number rounded to four decimal places.

Answer 1

0.0013 probability that at least 6 employees were over 50.

Explanation:

The employees were "chosen" to be dismissed without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x sucesses 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 sucesses.

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 problem:

8 employees dismissed means that
`n = 8`

Had 7 + 17 = 24 employees, which means that
`N = 24`

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

What is the probability that at least 6 employees were over 50?

6 or 7, so:

`P(X \geq 6) = P(X = 6) + P(X = 7)`.

In which

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

`P(X = 6) = h(6,24,8,7) = (C_(7,6)*C_(17,2))/(C_(24,8)) = 0.0013`

`P(X = 7) = h(7,24,8,7) = (C_(7,7)*C_(17,1))/(C_(24,8)) \approx 0`

`P(X \geq 6) = P(X = 6) + P(X = 7) = 0.0013 + 0 = 0.0013`

0.0013 probability that at least 6 employees were over 50.