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).
Need a fast expert's response?
Submit orderand get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Ask Your question
Comments
Leave a comment