Question

Certify Completion Icon Tries remaining:2 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

Answer:

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

Step-by-step explanation:

The employees are removed 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:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

7 + 18 = 25 employees, which means that $N = 25$

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

10 dismissed, which means that $n = 10$

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

This is P(X = 5). So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 5) = h(5,25,10,7) = \frac{C_{7,5}*C_{18,5}}{C_{25,10}} = 0.055$

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