Question

at a certain pizza parlor, 40% of the customers order a pizza containing onions, 33% of the customers order a pizza containing sausage, and 62% order a pizza containing onions or sausage (or both). find the probability that a customer chosen at random will order a pizza containing both onions and sausage. write your answer as a decimal (not as a percentage).

Explanation:

Step1: Recall the inclusion - exclusion principle for probability

The formula for the probability of the union of two events \(A\) and \(B\) is \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), where \(P(A\cup B)\) is the probability of \(A\) or \(B\) (or both), \(P(A)\) is the probability of \(A\), \(P(B)\) is the probability of \(B\), and \(P(A\cap B)\) is the probability of both \(A\) and \(B\).

Let \(A\) be the event that a customer orders a pizza with onions and \(B\) be the event that a customer orders a pizza with sausage. We know that \(P(A) = 0.4\), \(P(B)=0.33\), and \(P(A\cup B)=0.62\). We need to find \(P(A\cap B)\).

Step2: Rearrange the formula to solve for \(P(A\cap B)\)

From \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), we can rearrange it to \(P(A\cap B)=P(A)+P(B)-P(A\cup B)\).

Step3: Substitute the given values into the formula

Substitute \(P(A) = 0.4\), \(P(B)=0.33\), and \(P(A\cup B)=0.62\) into the formula:

\(P(A\cap B)=0.4 + 0.33-0.62\)

First, calculate \(0.4+0.33 = 0.73\). Then, calculate \(0.73 - 0.62=0.11\).

Answer:

\(0.11\)