Chapter 9.7, Problem 62E

a.

To determine

To find: the probability of event that all the children are boys.

Answer to Problem 62E

  $\frac{1}{16}$

Explanation of Solution

Concept Used:

Let A and B be two independent events, then

  $P\left(A\text{ and }B\right)=P\left(A\right)\times P\left(B\right)$

Let B be the event that all the children are boys

Probability of a child being a girl or a boy is same $\frac{1}{2}$ .

Also, the events that a child is girl or boy are independent events.

So, the probability that all the children are boys is,

  $\begin{array}{c}P\left(B\right)=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\\ =\frac{1}{16}\end{array}$

b.

To determine

To find: the probability of event that all the children the same sex.

Answer to Problem 62E

  $\frac{1}{8}$

Explanation of Solution

Concept Used:

Let A and B be two independent events, then

  $P\left(A\text{ and }B\right)=P\left(A\right)\times P\left(B\right)$

Let E be the event that all the children the same sex.

Probability of a child being a girl or a boy is same $\frac{1}{2}$ .

From part (a), the probability that all the children are boys is,

  $\begin{array}{c}P\left(B\right)=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\\ =\frac{1}{16}\end{array}$

Similarly, the probability that all the children are boys is,

  $\begin{array}{c}P\left(G\right)=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\\ =\frac{1}{16}\end{array}$

So, the probability that all the children are the same sex is,

  $\begin{array}{c}P\left(E\right)=P\left(\text{all girls}\right)+P\left(\text{all boys}\right)\\ =\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\\ =\frac{1}{16}+\frac{1}{16}\\ =\frac{1}{8}\end{array}$

c.

To determine

To find: the probability of event that there is at least one boy.

Answer to Problem 62E

  $\frac{15}{16}$

Explanation of Solution

Let E be the event that there is at least one boy.

Now, the even E is complement of the event G that all children are girls.

So,

  $P\left(E\right)=1-P\left(G\right)$

Probability of a child being a girl or a boy is same $\frac{1}{2}$ .

Also, the events that a child is girl or boy are independent events.

So, the probability that all the children are girls is,

  $\begin{array}{c}P\left(G\right)=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\\ =\frac{1}{16}\end{array}$

Thus, the probability that there is at least one boy is,

  $\begin{array}{c}P\left(E\right)=1-P\left(G\right)\\ =1-\frac{1}{16}\\ =\frac{15}{16}\end{array}$