Question

A firm offers routine physical examinations as part of a health service program for its employees. The exams showed that 22% of the employees needed corrective shoes, 29% needed major dental work, and 3% needed both corrective shoes and major dental work. What is the probability that an employee selected at random will need either corrective shoes or major dental work

Answer 1

Answer:

0.48 = 48% probability that an employee selected at random will need either corrective shoes or major dental work

Step-by-step explanation:

I am going to solve this question treating these events as Venn probabilities.

I am going to say that:

Event A: Employee needed corrective shoes.

Event B: Employee needed major dental work.

22% of the employees needed corrective shoes

This means that $P(A) = 0.22$

29% needed major dental work

This means that $P(B) = 0.29$

3% needed both corrective shoes and major dental work.

This means that $P(A \cap B) = 0.03$

What is the probability that an employee selected at random will need either corrective shoes or major dental work?

This is:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Replacing the values that we have:

$P(A \cup B) = 0.22 + 0.29 - 0.03 = 0.48$

0.48 = 48% probability that an employee selected at random will need either corrective shoes or major dental work