# Answer to Question #11231 in Statistics and Probability for Bonnie Jones

Question #11231

At a particular restaurant, 63% of all customers order an appetizer and 40% of all customers order dessert. If 76% of all customers order an appetizer or dessert (or both), what is the probability a randomly selected customer orders both an appetizer and dessert?

Expert's answer

In terms of probablity (A - appetizer oredered, B - desert ordered)

P(A)=63%

P(B)=40%

P(AorB)=76%.

Using the formula:

P(AorB)=P(A)+P(B)-P(A&B) we get P(A&B)=63%+40%-76%=27% (probability we were looking for).

P(A)=63%

P(B)=40%

P(AorB)=76%.

Using the formula:

P(AorB)=P(A)+P(B)-P(A&B) we get P(A&B)=63%+40%-76%=27% (probability we were looking for).

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