At a certain fast-food restaurant, 63% of customers order a chicken sandwich, 43% of customers order french fries, and 32% of customers order both a chicken sandwich and french fries. What is the probability that a randomly selected customer will order a chicken sandwich or french fries (or both items)?2 points0.650.690.740.83
Question
At a certain fast-food restaurant, 63% of customers order a chicken sandwich, 43% of customers order french fries, and 32% of customers order both a chicken sandwich and french fries. What is the probability that a randomly selected customer will order a chicken sandwich or french fries (or both items)?2 points0.650.690.740.83
Solution
To solve this problem, we use the formula for the probability of the union of two events: P(A U B) = P(A) + P(B) - P(A ∩ B).
Here, ordering a chicken sandwich is event A, and ordering french fries is event B. The problem gives us the probabilities of these events, as well as the probability of both events occurring (the intersection of A and B).
So, we have:
P(A) = 0.63 (the probability of ordering a chicken sandwich) P(B) = 0.43 (the probability of ordering french fries) P(A ∩ B) = 0.32 (the probability of ordering both)
Substituting these values into the formula gives us:
P(A U B) = 0.63 + 0.43 - 0.32 = 0.74
So, the probability that a randomly selected customer will order a chicken sandwich or french fries (or both) is 0.74, or 74%. Therefore, the correct answer is 0.74.
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