Question

A bag contains $4$ red and $3$ black balls. A second bag contains $2$ red and $4$ black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.

• A. $\frac{23}{42}$
• B. $\frac{19}{42}$
• C. $\frac{7}{32}$
• D. $\frac{16}{39}$

Answer 1

The correct option is B

$\frac{19}{42}$

Explanation of the correct option:

Finding the required probability:

Let $A$ be the event of selecting one out of two bags.

$\Rightarrow P\left(A\right)=\frac{1}{2}$

Let $B$ be the event of drawing a red ball from the first bag.

$\Rightarrow P\left(B\right)=\frac{4}{7}$ (out of total $7$ balls, there are $4$ red balls)

Let $C$ be the event of drawing a red ball from the second bag.

$\Rightarrow P\left(C\right)=\frac{2}{6}$(out of total $6$ balls, there are $2$ red balls)

We have to find the probability that the ball is drawn from the selected bag is red. It can be possible in either of the following ways -

Selecting the first bag and then drawing a red ball from it $=P\left(A\right)\times P\left(B\right)$

Selecting the second bag and then drawing a red ball from it $=P\left(A\right)\times P\left(C\right)$

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

$\begin{array}{rcl}P\left(A\right)\times P\left(B\right)+P\left(A\right)\times P\left(C\right)& =& \left[\frac{1}{2}\times \frac{4}{7}\right]+\left[\frac{1}{2}\times \frac{2}{6}\right]\\ & =& \frac{2}{7}+\frac{1}{6}\\ & =& \frac{12+7}{42}\\ & =& \frac{19}{42}\end{array}$

Hence, the correct answer is option B.