Chapter 11.6, Problem 32PPSE

Solution

To identify:The correct optionthe probability that the student is a girl or has a last name that begins with S A homeroom class consists of 6 boys whose last name begins with S. 8 boys whose last name begins with 1, 4girls whose last name begins with S, and 11 girls whose last name begins with TA student is chosen at random from the class.

Options (A) $\frac{18}{29}$

Options (B) $\frac{21}{29}$

Options (C) $\frac{23}{29}$

Options (D) $\frac{25}{29}$

Option (C) is correct option is $\frac{21}{29}$

Given information:

Class consists of 6 boys whose last name begins with S. 8 boys whose last name begins with 1, 4girls whose last name begins with S, and 11 girls whose last name begins with TA student is chosen at random from the class.

Explanation:

Consider the given information.

Total number of students in the class is

  $6+8+4-11=29$

Let G shows the event that the student will be a girl. Total number of girl in the class is $4+11=15$ So the probability that the selected student will be girl is

  $P\left(G\right)=\frac{15}{29}$

Let S shows the event that selected student's last name begin with S. Total number of students in the class whose last name begin with S is $4+6=10.$ So the probability of event S is

  $P\left(S\right)=\frac{10}{29}$

There are 4 girls whose last name begins with S so

  $P(GandS)=\frac{4}{29}$

So the probability that the student is a girl or has a last name begins with S will be

  $\begin{array}{c}P\left(GorS\right)\text{= }P\left(G\right)+P\left(S\right)-P\left(GandS\right)\\ =\frac{15}{29}+\frac{10}{29}-\frac{4}{29}\\ =\frac{15+10-4}{29}\\ =\frac{21}{29}\end{array}$

So, the probability that the student is a girl or has a last name begins with Sis $\frac{21}{29}$

So, option B is correct.