Question

In a clothing store, 65% of the customers buy a shirt, 30% of the customers
buy a pair of pants, and 20% of the customers buy both a shirt and a pair of
pants.
If a customer is chosen at random, what is the probability that he or she buys
a shirt or a pair of pants?

Answer 1

Answer:

75%

Step-by-step explanation:

List out known probabilities

$\Pr(shirt) = 0.65 = \frac{65}{100} = \frac{13}{20}\\ \Pr(pants) = 0.40 = \frac{30}{100} = \frac{6}{20}\\\Pr(shirt \cap pants) = 0.20 = \frac{20}{100}=\frac{1}{5}$

recall equation

$\Pr(A \cup B)=\Pr(A)+\Pr(B)-\Pr(A \cap B)$

Plug in values

$\Pr(shirt \cup pants)=\frac{13}{20}+\frac{6}{20}-\frac{1}{5}\\\\therefore \Pr(shirt \cup pants) = \frac{15}{20} =\frac{3}{4} = 75\%$