Question

A town recently dismissed 5 employees in order to meet their new budget reductions. The town had 4 employees over 50 years of age and 16 under 50. If the dismissed employees were selected at random, what is the probability that no more than 1 employee was over 50

Answer 1

Answer:

0.7513 = 75.13% probability that no more than 1 employee was over 50

Step-by-step explanation:

The employees are chosen 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:

4 + 16 = 20 employees, which means that $N = 20$

4 over 50, which means that $k = 4$

5 were dismissed, which means that $n = 5$

What is the probability that no more than 1 employee was over 50?

Probability of at most one over 50, which is:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

In which

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

$P(X = 0) = h(0,20,5,4) = \frac{C_{4,0}*C_{16,5}}{C_{20,5}} = 0.2817$

$P(X = 1) = h(1,20,5,4) = \frac{C_{4,1}*C_{16,4}}{C_{20,5}} = 0.4696$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.2817 + 0.4696 = 0.7513$

0.7513 = 75.13% probability that no more than 1 employee was over 50