Chapter 1.4, Problem 53HP

To determine

To find: the probability that the student will be either an 11^th grader or male.

Answer to Problem 53HP

The probability that the student will be either an 11^th grader or male is $\frac{5}{8}$ .

Explanation of Solution

Given information:

Two male and 2 female students from given of the 9^th, 10^th ,11^th, and 12^th grades comprise the student council, If a student Council representative is chosen at random to attend a board meeting.

Calculation:

Define the events,

Let $A$ and $B$ be selecting a 11^th grade student and a male student respectively.

Since the events $A$ and $B$ can occur at the same time, they are not mutually exclusive events.

  $P\left(AorB\right)=P\left(A\right)+P\left(B\right)-P\left(AandB\right)$

Recall the probability of events that are not mutually exclusive.

If two events A and B are not mutually exclusive, then the probability that either A or B occurs is the sum of their probabilities decreased by the probability of both occurring.

Find the probability of event A.

  $P\left(A\right)=\frac{4}{16}=\frac{1}{4}$

Find the probability of event B.

  $P\left(B\right)=\frac{8}{16}=\frac{1}{2}$

Find the probability of event A and B.

  $P\left(AandB\right)=P\left(A\right)+P\left(B\right)=\frac{1}{8}$

Find the probability of event A or B.

  $\begin{array}{c}P\left(AorB\right)=P\left(A\right)+P\left(B\right)-P\left(AandB\right)\\ =\frac{1}{4}+\frac{1}{2}+\left(-\frac{1}{8}\right)\\ =\frac{5}{8}\end{array}$

Therefore, the probability that the student will be either an 11^th grader or male is $\frac{5}{8}$ .