A retail shop accepts only cash or checks. Suppose that 54% of its customers carry cash, 51% carry checks, and 71% carry cash or checks (or both). What is the probability that a randomly chosen customer at the shop is carrying both cash and checks?

Pr(Cash or check)=Pr(cash)+Pr(check)-Pr(cash AND check)

pr(Cash AND check)=(.51+.54-.71)

http://www.pindling.org/Math/Statistics/Textbook/Chapter4_Probability/compound_events.htm cash, check are not mutually exclusive

A retail shop accepts only cash or checks. Suppose that

46

% of its customers carry cash,
37

% carry checks, and
74

% carry cash or checks (or both). What is the probability that a randomly chosen customer at the shop is carrying both cash and checks?

Well, it sounds like this shop is really cashing in on its customers! Anyway, to find the probability that a customer is carrying both cash and checks, we can use a simple equation. We know that 71% of customers carry either cash or checks (or both), so that means 71% of customers are in the "cash or checks" group. Now, if 54% of customers carry cash and 51% carry checks, that means that 54% + 51% = 105% of customers carry cash and/or checks.

Since we can't have more than 100% of customers carrying cash and/or checks (unless they've invented some new mathematical concept where customers can carry more than 100% of things, which would be pretty impressive), we need to subtract the extra 5% from the total. So, the probability that a randomly chosen customer is carrying both cash and checks is 71% - 5% = 66%.

So, it's pretty likely that the customer has both cash and checks! They're just covering all their financial bases, you know?

To find the probability that a randomly chosen customer is carrying both cash and checks, we can use the principle of inclusion-exclusion.

First, let's assign some variables:
Let C represent the event of carrying cash.
Let Ch represent the event of carrying checks.

We are given the following probabilities:
P(C) = 0.54
P(Ch) = 0.51
P(C or Ch) = 0.71

Now, using the principle of inclusion-exclusion, we can find the probability of carrying both cash and checks.

P(C and Ch) = P(C) + P(Ch) - P(C or Ch)

P(C and Ch) = 0.54 + 0.51 - 0.71

P(C and Ch) = 0.54 + 0.51 - 0.71

P(C and Ch) = 0.54 + 0.51 - 0.71

P(C and Ch) = 1.05 - 0.71

P(C and Ch) = 0.34

Therefore, the probability that a randomly chosen customer at the shop is carrying both cash and checks is 0.34.