Chapter 0.6, Problem 9E

To determine

To find: the probability of getting third marble as red given that first two are red and blue that cannot be replaced.

Answer to Problem 9E

The probability of $P\left(\text{red/red and blue}\right)$ is $\frac{5}{17}$ .

Explanation of Solution

Given:

A bag contains $8$ blue marbles, $6$ red marbles and $5$ green marbles.

Concept used:

The number of ways of arranging $n$ item in order $=n!$ .

The probability formula is used to compute the probability of an event to occur as:

  $P\left(A\right)=\frac{\text{Number of favorable outcome}}{\text{Total Number of favorable outcomes}}$ .

Calculation:

Consider a bag contains $8$ blue marbles, $6$ red marbles and $5$ green marbles. Three marbles are drawn at a time. The probability of second marble is green and the first marble is blue and not replaced. Consider the following probability:

  $P\left(\text{red/red and blue}\right)$:

Use the conditional probability for the above probability is as follows:

  $P\left(\text{red/red and blue}\right)=\frac{P\left(\text{red and }\left(\text{red and blue}\right)\right)}{P\left(\text{red and blue}\right)}$ .

Here,

  $\begin{array}{l}P\left(\text{red and blue}\right)=\frac{6}{19}\times \frac{8}{18}.\\ P\left(\text{red and }\left(\text{red and blue}\right)\right)=\frac{6}{19}\times \frac{5}{18}\times \frac{8}{17}.\\ \Rightarrow \frac{240}{5814}.\end{array}$

Substituting the value and simplify it as follows:

  $\begin{array}{l}P\left(\text{red/red and blue}\right)=\frac{P\left(\text{red and }\left(\text{red and blue}\right)\right)}{P\left(\text{red and blue}\right)}.\\ \Rightarrow \frac{\frac{240}{5814}}{\frac{48}{342}}.\\ \Rightarrow \frac{5}{17}.\end{array}$

Hence, the probability of $P\left(\text{red/red and blue}\right)$ is $\frac{5}{17}$ .