(A)$\large\frac{1}{10}$

(B) $\large\frac{3}{5}$

(C) $\large\frac{2}{5}$

(D) $\large\frac{3}{10}$

- Baye's Theorem:Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability $P(E_i|A)$ for any event A associated with $E_i$ as follows: \(\;P(E_i/A)\)=\(\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}\)

Let $A$ be the event of drawing $2$ white balls from the bag containing 4 balls.

The remaining $2$ balls of the bag has three options.

Let $E_1$ be the event that the remaining $2$ balls of the bag are not white.

Let $E_2$ be the event that the remaining $2$ balls of the bag are one white and one not white.

Let $E_2$ be the event that the remaining $2$ balls of the bag are white.

All the three options are equally likely.

$\therefore\:P(E_1) = P(E_2)=P(E_3)= \large\frac{1}{3}$

$P(A/E_1)=P($ drawing $2$ white balls from the bag containing $2$ white and $2$ non white balls.)

$=\large\frac{^2C_2}{^4C_2}=\frac{1}{6}$

$P(A/E_2)=P($ drawing $2$ white balls from the bag containing $3$ white and $1$ non white balls.)

$=\large\frac{^3C_2}{^4C_2}=\frac{3}{6}=\frac{1}{2}$

$P(A/E_3)=P($ drawing $2$ white balls from the bag containing $4$ white balls)

$=1$

We have to find the probability that the remaining balls are $1$ white and $1$ non white.

$i.e.,$ We have to find $P(E_2/A)$

According to Baye's theorem, $P(E_2|A) = \large\frac{P(E_2)(P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)(PA|E_3)}$

$P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)(PA|E_3)$

$=\large\frac{1}{3}$$\times\large\frac{1}{6}$$+\large\frac{1}{3}$$\times\large\frac{1}{2}$$+\large\frac{1}{3}$$\times 1=\large\frac{1}{3}$$(\large\frac{1}{6}$$+\large\frac{1}{2}$$+1)$$=\large\frac{1}{3}$$\times\large\frac{5}{3}$

$\therefore\:P(E_2/A)=\large\frac{\large\frac{1}{3}\times\frac{1}{2}}{\large\frac{1}{3}\times \frac{5}{3}}=\frac{3}{10}$

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