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:

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

Step-by-step 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 [tex]P(A) = 0.14[/tex]

21% needed major dental work

This means that [tex]P(B) = 0.21[/tex]

3% needed both corrective shoes and major dental work.

This means that [tex]P(A \cap B) = 0.03[/tex]

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

This is:

[tex]P(A \cup B) = P(A) + P(B) - P(A \cap B)[/tex]

So, from the values given by the exercise:

[tex]P(A \cup B) = 0.14 + 0.21 - 0.03 = 0.32[/tex]

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

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