The manager of a Drug Store knows that 50% of the customers entering the store buy prescription drugs, 65% buy over-the-counter drugs, and 18% buy both types of drugs. What is the probability that a randomly selected customer will buy at least one of these two types of drugs?
A).5
B).85
C).97
D)1.0
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Prob(A OR B) = Prob(A) + Prob(B) - Prob(A AND B)
= .5 + .65 - .18
= .97
looks like C)
To find the probability that a randomly selected customer will buy at least one of the two types of drugs, we need to subtract the probability of the customer not buying any drugs from 1.
Let's denote:
P(Prescription drugs) = 0.50
P(Over-the-counter drugs) = 0.65
P(Both prescription and over-the-counter drugs) = 0.18
To find the probability of not buying any drugs, we need to find the complement of the event that the customer buys at least one of the two types of drugs, which is 1 - P(Buys at least one of the two types of drugs).
To calculate P(Buys at least one of the two types of drugs), we can use the principle of inclusion-exclusion. This principle states that the probability of the union of two events (A and B) is equal to the sum of their individual probabilities, minus the probability of their intersection.
Probability of buying at least one of the two drugs = P(Prescription drugs) + P(Over-the-counter drugs) - P(Both prescription and over-the-counter drugs)
= 0.50 + 0.65 - 0.18
= 0.97
Therefore, the probability that a randomly selected customer will buy at least one of the two types of drugs is 0.97.
The correct answer is option C) 0.97.