Question

The mathematics department of a college has 8 male professors , 11 female professors 5 male teaching assistants , and 5 female teaching assistants . If a person is selected at random from the group , find the probability that the selected person is a professor or a male

Answer 1

Answer:

The probability that the selected person is a professor or a male is;

$\frac{24}{29}$

Explanation:

Given that the mathematics department of a college has 8 male professors, 11 female professors 5 male teaching assistants, and 5 female teaching assistants.

let A represent professors and B represent males.

the probability that the selected person is a professor or a male is;

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

Solving or the probability that the selected person is a professor;

$P(A)=\frac{n(A)}{n(T)}=\frac{\text{number of professors}}{\text{total number of persons}}$$P(A)=\frac{8+11}{8+11+5+5}=\frac{19}{29}$

The probability that the selected person is a male is;

$P(B)=\frac{n(B)}{n(T)}=\frac{\text{ number of males }}{\text{ Total number of persons}}$$P(B)=\frac{8+5}{8+11+5+5}=\frac{13}{29}$

Then the probability that the selected person is a male and a professor;

$P(A\cap B)=\frac{n(A\cap B)}{n(T)}=\frac{\text{ number of male professors}}{\text{total number of persons}}$$P(A\cap B)=\frac{8}{8+11+5+5}=\frac{8}{29}$

We can now substitute to get the probability that the selected person is a professor or a male;

$\begin{gathered} P(A\cup B)=P(A)+P(B)-P(A\cap B) \\ P(A\cup B)=\frac{19}{29}+\frac{13}{29}-\frac{8}{29} \\ P(A\cup B)=\frac{19+13-8}{29} \\ P(A\cup B)=\frac{24}{29} \end{gathered}$

Therefore, the probability that the selected person is a professor or a male is;

$\frac{24}{29}$