Chapter 13.5, Problem 17E

To determine

To calculate: The probability that girls are separated provided that a girl is at the end of line.

Answer to Problem 17E

The probability that girls are separated provided that a girl is at the end of line is $\frac{3}{5}$ .

Explanation of Solution

Given information:

In a line two boys and two girls are randomly lined up.

Formula used:

Probability of an event E is $P\left(\text{E}\right)=\frac{\text{number of outcome}}{\text{total number of outcome}}$ .

Given an event Q, the conditional probability of event A is $P\left(A|Q\right)=\frac{P\left(A\text{ and Q}\right)}{P\left(Q\right)}$ where $P\left(Q\right)\ne 0$ .

Calculation:

Consider the provided information that in a line two boys and two girls are randomly lined up.

The total number of ways in which 4 people can be lined up is $4!$ as there are 4 ways to fill the first seat, 3 ways to fill the second seat, two ways to fill the second seat and 1 way to fill the last one.

Let A be an event that girls are separated and B be an event that girl is at the end.

Denote two girls as g and G and two boys as b and B.

The following cases are when girls are separated and a girl is at the end,

$\begin{array}{l}g,b,G,B\\ G,b,g,B\\ g,b,B,G\\ G,b,B,g\end{array}$

And,

$\begin{array}{c}b,g,B,G\\ b,G,B,g\end{array}$

There are total $6\cdot 2=12$ ways.

The combinations when girl is not at end are,

$\begin{array}{l}b,g,G,B\\ b,G,g,B\end{array}$

There are total $2\cdot 2=4$ ways.

Recall that probability of an event E is $P\left(\text{E}\right)=\frac{\text{number of outcome}}{\text{total number of outcome}}$ .

Probability that girl is at end,

  $\begin{array}{c}P\left(B\right)=1-\frac{4}{4!}\\ =1-\frac{4}{24}\\ =\frac{20}{24}\end{array}$

Probability that girls are separated and a girl is at the end,

  $P\left(A\text{ and B}\right)=\frac{12}{24}$

Recall that given an event Q, the conditional probability of event A is $P\left(A|Q\right)=\frac{P\left(A\text{ and Q}\right)}{P\left(Q\right)}$ where $P\left(Q\right)\ne 0$ .

The probability that girls are separated provided that a girl is at the end of line.

  $\begin{array}{c}P\left(A|B\right)=\frac{\frac{12}{24}}{\frac{20}{24}}\\ =\frac{12}{20}\\ =\frac{3}{5}\end{array}$

Thus, the probability that girls are separated provided that a girl is at the end of line is $\frac{3}{5}$ .