Chapter 11.3, Problem 39PPSE

Solution

To identify:The probability one of the marble you select is blue and the other is red.A jar contains four blue marbles and two red marbles. Suppose you choose a marble at random, and do not replace it. Then you choose a second marble.

The probability is $\left(blue,thenred\right)+\left(red,then\text{ blue}\right)=\frac{8}{15}$

Given information:

A jar contains four blue marbles and two red marbles.

Explanation:

Consider the information.

Total marbles are $4+2=6$

So, probability of selecting red marble is:

  $\begin{array}{l}P\left(red\right)=\frac{2}{6}\\ P\left(red\right)=\frac{1}{3}\end{array}$

Since the marble is not replaced and have only 5 marbles left, 4 blue and 1 red.

So probability of selecting red marble is

  $P\left(blue\right)=\frac{4}{5}$

So,

  $\begin{array}{c}P\left(red,then\text{ blue}\right)=P\left(red\right)\cdot P\left(then\text{ blue}\right)\\ =\frac{1}{3}\cdot \frac{4}{5}\\ =\frac{4}{15}\end{array}$

Hence $P\left(red,then\text{ blue}\right)\text{ =}\frac{4}{15}$

Similarly probability of selecting blue marble first is

  $\begin{array}{l}P\left(red\right)=\frac{4}{6}\\ P\left(red\right)=\frac{2}{3}\end{array}$

Since the marble is not replaced and have only 5 marbles left, 3 blue and 2 red.

So probability of selecting red marble is

  $P\left(red\right)=\frac{2}{5}$

So,

  $\begin{array}{c}P\left(blue,then\text{ red}\right)=P\left(blue\right)\cdot P\left(then\text{ red}\right)\\ =\frac{2}{3}\cdot \frac{2}{5}\\ =\frac{4}{15}\end{array}$

Hence $P\left(red,then\text{ blue}\right)\text{ =}\frac{4}{15}$

So the total probability of selecting one blue and other red marble is

  $\begin{array}{c}P=\left(blue,thenred\right)+\left(red,then\text{ blue}\right)\\ P=\frac{4}{15}+\frac{4}{15}\\ P=\frac{8}{15}\end{array}$

Hence $P=\frac{8}{15}$