Question

at a certain fast - food restaurant, 68% of customers order a chicken sandwich, 66% of customers order french fries, and 55% 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)? write your answer as a decimal (not as a percentage).

Explanation:

Step1: Identify the probability formula

Use the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Let $A$ be the event of ordering a chicken - sandwich and $B$ be the event of ordering french fries.

Step2: Convert percentages to decimals

$P(A) = 0.68$, $P(B)=0.66$, and $P(A\cap B)=0.55$.

Step3: Calculate the probability

$P(A\cup B)=0.68 + 0.66-0.55$.
$P(A\cup B)=0.79$.

Answer:

$0.79$