Routine physical examinations are conducted annually as part of a health service program for the employees. It was discovered 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?
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A. B. C. D. E.Explanation
8% + 15% - 3% = 20%
To solve this problem, we can use the principle of inclusion-exclusion. Let's define the following events:
Event A: Employee needs corrective shoes. Event B: Employee needs major dental work.
We are given the following probabilities:
P(A) = 0.08 (8% of employees need corrective shoes) P(B) = 0.15 (15% of employees need major dental work) P(A ∩ B) = 0.03 (3% of employees need both corrective shoes and major dental work)
To find the probability that an employee selected at random will need either corrective shoes or major dental work, we need to find P(A ∪ B), which represents the probability of the union of events A and B.
We can use the formula:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
Substituting the given probabilities:
P(A ∪ B) = 0.08 + 0.15 - 0.03 = 0.23
Therefore, the probability that an employee selected at random will need either corrective shoes or major dental work is 0.23.
None of the provided answer choices matches the calculated probability.