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 chosen to be white is

विकल्प

•  2/15

• 7/15

• 8/15

• 14/15

   

Answer 1

8/15

A white ball can be drawn in two mutually exclusive ways:
(I) Selecting bag X and then drawing a white ball from it.
(II) Selecting bag Y and then drawing a white ball from it.

Let E₁, E₂ and A be three events as defined below:
E₁ = Selecting bag X
E₂ = Selecting bag Y
A = Drawing a white ball

We know that one bag is selected randomly.

\[\therefore P\left( E_1 \right) = \frac{1}{2} \]
\[ P\left( E_2 \right) = \frac{1}{2}\]
\[ P\left( A/ E_1 \right) = \frac{2}{5}\]
\[P\left( A/ E_2 \right) = \frac{4}{6} = \frac{2}{3}\]
\[\text{ Using the law of total probability, we get} \]
\[\text{ Required probability } = P\left( A \right) = P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)\]
\[ = \frac{1}{2} \times \frac{2}{5} + \frac{1}{2} \times \frac{2}{3}\]
\[ = \frac{1}{5} + \frac{1}{3}\]
\[ = \frac{3 + 5}{15} = \frac{8}{15}\]