Question

at a certain fast-food restaurant, 59% of customers order a chicken sandwich, 49% of customers order french fries, and 33% 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: Recall the formula for the probability of the union of two events

The formula for \( P(A \cup B) \) (the probability that event \( A \) or event \( B \) occurs) is \( P(A \cup B)=P(A)+P(B)-P(A \cap B) \), where \( P(A) \) is the probability of event \( A \), \( P(B) \) is the probability of event \( B \), and \( P(A \cap B) \) is the probability of both \( A \) and \( B \) occurring.

Let \( A \) be the event that a customer orders a chicken sandwich and \( B \) be the event that a customer orders french fries. We know that \( P(A) = 0.59 \) (since 59% = 0.59), \( P(B)=0.49 \) (since 49% = 0.49), and \( P(A \cap B) = 0.33 \) (since 33% = 0.33).

Step2: Substitute the values into the formula

Substitute the given values into the formula \( P(A \cup B)=P(A)+P(B)-P(A \cap B) \):

\( P(A \cup B)=0.59 + 0.49- 0.33 \)

First, add \( 0.59 \) and \( 0.49 \): \( 0.59+0.49 = 1.08 \)

Then, subtract \( 0.33 \) from the result: \( 1.08 - 0.33=0.75 \)

Answer:

\( 0.75 \)