Question

a restaurant has a total of 60 tables. of those tables, 38 are round and 13 are located by the window. there are 6 round tables by the window. if tables are randomly assigned to customers, what is the probability that a customer will be seated at a round table or by the window?
○ a. $\frac{47}{60}$
○ b. $\frac{41}{60}$
○ c. $\frac{38}{60}$
○ d. $\frac{45}{60}$

Explanation:

Step1: Recall probability formula for union

The formula for $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Let $A$ be the event of being seated at a round - table and $B$ be the event of being seated by the window.

Step2: Calculate $P(A)$, $P(B)$ and $P(A\cap B)$

The total number of tables $n = 60$. The number of round tables $n(A)=38$, so $P(A)=\frac{38}{60}$. The number of tables by the window $n(B)=13$, so $P(B)=\frac{13}{60}$. The number of round tables by the window $n(A\cap B)=6$, so $P(A\cap B)=\frac{6}{60}$.

Step3: Substitute values into the formula

$P(A\cup B)=\frac{38}{60}+\frac{13}{60}-\frac{6}{60}=\frac{38 + 13-6}{60}=\frac{45}{60}$.

Answer:

D. $\frac{45}{60}$