Question

a retail shop accepts only cash or checks. suppose that 46% of its customers carry cash, 42% carry checks, and 64% carry cash or checks (or both). what is the probability that a randomly chosen customer at the shop is carrying both cash and checks? write your answer as a decimal (not as a percentage).

Explanation:

Step1: Recall the formula for the union of two events

Let $A$ be the event that a customer carries cash and $B$ be the event that a customer carries checks. The formula for $P(A\cup B)$ is $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.

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

We can rewrite the formula as $P(A\cap B)=P(A)+P(B)-P(A\cup B)$.

Step3: Substitute the given values

We know that $P(A) = 0.46$, $P(B)=0.42$ and $P(A\cup B)=0.64$. Substituting these values into the formula, we get $P(A\cap B)=0.46 + 0.42- 0.64$.

Step4: Calculate the result

$0.46+0.42 - 0.64=0.24$.

Answer:

$0.24$