Question

A bag $X$ contains $2$ white and $3$ black balls and another bag $Y$ contains $4$ white and $2$ black balls. One bag is selected at random and a ball is drawn from it. Then the probability for the ball chosen be white is

• A. $\frac{2}{15}$
• B. $\frac{7}{15}$
• C. $\frac{8}{15}$
• D. $\frac{14}{15}$

Answer 1

The correct option is C

$\frac{8}{15}$

Explanation for the correct option:

Step 1. Find the probability of getting white ball:

Given, Total balls in bag $“X”$$=5$

Total balls in bag $“Y”$$=6$

Let $E_x$ be the event that the ball is drawn from bag $“X”$ and $E_y$ be the event that the ball is drawn from bag $“Y”$.

Let $E_W$ be the event of the ball drawn is white.

As we know both the bags are equally likely to be selected, we have

$P\left(E_x\right)=\frac{1}{2}$ and $P\left(E_y\right)=\frac{1}{2}$

Now, probability of getting white ball from bag $“X”$ is, $P\left(\frac{E_w}{E_x}\right)=\frac{2}{5}$, $\left[\mathbf{\because }\mathit{P}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{\left(}\mathbf{E}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}\right]$

Probability of getting white ball from bag $“Y”$ is, $\begin{array}{rcl}P\left(\frac{E_w}{E_y}\right)& =& \frac{4}{6}\end{array}$

$\begin{array}{rcl}& =& \frac{2}{3}\end{array}$

Step 2. The probability of getting white ball from either of the bag is, $\mathit{P}\mathbf{\left(}{\mathit{E}}_{\mathbf{w}}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathit{P}\mathbf{\left(}{\mathit{E}}_{\mathbf{x}}\mathbf{\right)}\mathbf{.}\mathit{P}\left(\frac{E_w}{E_x}\right)\mathbf{}\mathbf{+}\mathit{P}\mathbf{\left(}{\mathit{E}}_{\mathbf{y}}\mathbf{\right)}\mathbf{.}\mathit{P}\left(\frac{E_w}{E_y}\right)$

$$\begin{gathered} =\frac{1}{2}\times \frac{2}{5}+\frac{1}{2}\times \frac{2}{3} \\ =\frac{2}{10}+\frac{2}{6} \\ =\frac{6+10}{30} \\ =\frac{16}{30} \\ =\frac{\mathbf{8}}{\mathbf{15}} \end{gathered}$$

Hence, Option ‘C’ is Correct.