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At a certain fast-food restaurant, 65% of customers order a chicken sandwich, 44% of customers order french fries, and 76% of customers order a chicken sandwich or french fries (or both items). What is the probability that a randomly selected customer will order both a chicken sandwich and french fries? Write your answer as a decimal (not as a percentage). (If necessary, consult a list of formulas.)

User Karin
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1 Answer

6 votes

Answer: 0.33

Explanation:

Let C be the event of ordering chicken sandwich.

F be the event of ordering french fries.

As per given , we have

P(C)=0.65

P(F)=0.44

P(C∩F)=0.76

Formula we use here,
P(A\cap B)=P(A)+P(B)-P(A\cup B)

Then, For given situation, we have


P(C\cap F)=P(C)+P(F)-P(C\cup F)

i.e.
P(C\cap F)=0.65+0.44-0.76=0.33

Hence, the probability that a randomly selected customer will order both a chicken sandwich and french fries=0.33

User Cprakashagr
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