Question

At a certain fast-food restaurant, 69 % of customers order a chicken sandwich, 46 % of customers order french fries, and 36 % 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)?

Answer 1

We have events:

A - customer orders a chicken sandwich
B - customer orders french fries


and probabilities:

$P(A)=0.69\\\\P(B)=0.46\\\\P(A\cap B)=0.36$

so:

$P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.69+0.46-0.36=\boxed{0.79}$