A firm offers routine physical examinations as part of a health service program for its employees. The exams showed that 8% of the employees needed corrective shoes, 15% 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?
A. 0.20
B. 0.25
C. 0.50
D. 1.00

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JeanaShupp JeanaShupp · Answer 1 · 2019-09-27T06:58:45+02:00

Answer: A. 0.20

Step-by-step explanation:

Let A be the event of employees needed corrective shoes and B be the event that they needed major dental work .

We are given that : [tex]P(A)=0.08\ ;\ P(B)=0.15\ ;\ P(A\cap B)=0.03[/tex]

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

Then, [tex]P(A\cup B)=0.08+0.15-0.03= 0.20[/tex]

Hence, the probability that an employee selected at random will need either corrective shoes or major dental work : [tex]P(A\cup B)= 0.20[/tex]

hence, the correct option is (A).

MrRoyal MrRoyal · Answer 2 · 2022-04-07T11:12:24+02:00

The probability that an employee selected at random will need either corrective shoes or major dental work is 0.20

The given parameters are:

P(corrective shoes) = 8% ,

P(major dental work) = 15%

P(Both) = 3%

The probability that an employee will need either of the two is calculated as:

P(Either) = P(corrective shoes) + P(major dental work) - P(Both)

This gives

P(Either) = 8% + 15% - 3%

Evaluate the expression

P(Either) = 0.20

Hence, the probability that an employee will need either of the two is 0.08