Question

If from each of the three boxes containing $3$ blue and $1$ red balls, $2$ blue and $2$ red balls, $1$ blue and $3$ red balls, one ball is drawn at random, then the probability that $2$ blue and $1$ red ball will be drawn is :

• A. $\frac{13}{32}$
• B. $\frac{27}{32}$
• C. $\frac{19}{32}$
• D. None of these

Answer 1

The correct option is A

$\frac{13}{32}$

Explanation of the correct option:

Finding the required probability:

Let $A$ be the event of drawing $1$ red ball and $B$ be the event of drawing $1$ blue ball from the first bag .

$\Rightarrow P\left(A\right)=\frac{1}{4}$ and $P\left(B\right)=\frac{3}{4}$ (out of total $4$ balls, there is $1$ red and $3$ blue balls)

Let $C$ be the event of drawing $1$ red ball and $D$ be the event of drawing $1$ blue ball from the second bag .

$\Rightarrow P\left(C\right)=\frac{2}{4}$ and $P\left(D\right)=\frac{2}{4}$ (out of total $4$ balls, there are $2$ red and $2$ blue balls)

Let $E$ be the event of drawing $1$ red ball and $F$ be the event of drawing $1$ blue ball from the third bag .

$\Rightarrow P\left(E\right)=\frac{3}{4}$ and $P\left(F\right)=\frac{1}{4}$ (out of total $4$ balls, there are $3$ red and $1$ blue ball)

We have to find the probability that $2$ blue and $1$ red ball will be drawn. It can be possible in either of the following ways -

Drawing $1$ blue ball from first bag, $1$ blue ball from second bag and $1$ red ball from third bag $=P\left(B\right)\times P\left(D\right)\times P\left(E\right)$

Drawing $1$ red ball from first bag, $1$ blue ball from second bag and $1$ blue ball from third bag $=P\left(A\right)\times P\left(D\right)\times P\left(F\right)$

Drawing $1$ blue ball from first bag, $1$ red ball from second bag and $1$ blue ball from third bag $=P\left(B\right)\times P\left(C\right)\times P\left(F\right)$

Therefore, the required probability is $=P\left(B\right)\times P\left(D\right)\times P\left(E\right)+P\left(A\right)\times P\left(D\right)\times P\left(F\right)+P\left(B\right)\times P\left(C\right)\times P\left(F\right)$

$\begin{array}{rcl}P\left(B\right)\times P\left(D\right)\times P\left(E\right)+P\left(A\right)\times P\left(D\right)\times P\left(F\right)+P\left(B\right)\times P\left(C\right)\times P\left(F\right)& =& \left[\frac{3}{4}\times \frac{2}{4}\times \frac{3}{4}\right]+\left[\frac{1}{4}\times \frac{2}{4}\times \frac{1}{4}\right]+\left[\frac{3}{4}\times \frac{2}{4}\times \frac{1}{4}\right]\\ & =& \frac{18}{64}+\frac{2}{64}+\frac{6}{64}\\ & =& \frac{26}{64}\\ & =& \frac{13}{32}\end{array}$

Hence, the correct answer is option A.