Question

A firm offers routine physical examinations as part of a health service program for its employees. The exams showed that 14% of the employees needed corrective shoes, 21% 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

0.32 = 32% probability that an employee selected at random will need either corrective shoes or major dental work.

Explanation:

We solve this question treating these events as Venn probabilities.

I am going to say that:

Event A: Needing corrective shoes.

Event B: Needing major dental work.

14% of the employees needed corrective shoes

This means that
`P(A) = 0.14`

21% needed major dental work

This means that
`P(B) = 0.21`

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)`

So, from the values given by the exercise:

`P(A \cup B) = 0.14 + 0.21 - 0.03 = 0.32`

0.32 = 32% probability that an employee selected at random will need either corrective shoes or major dental work.