Let

• $T_R$ = event that the transferred ball from Bag I to Bag II is red,
• $T_B$ = event that the transferred ball is black,
• $T_W$ = event that the transferred ball is white,
• $D_B$ = event that the ball drawn from Bag II is black.

We need to find $P(T_R\mid D_B).$

We use Bayes' theorem: $P(T_R\mid D_B)=\frac{P(T_R)P(D_B\mid T_R)}{P(D_B)}.$

1. Probabilities of transfer from Bag I

Bag I has $3$ red, $4$ black, $3$ white balls, total $10$ balls.

So, $P(T_R)=\frac{3}{10},\qquad P(T_B)=\frac{4}{10}=\frac{2}{5},\qquad P(T_W)=\frac{3}{10}.$

2. Probability of drawing a black ball from Bag II in each case

Initially, Bag II has $2$ red, $5$ black, $2$ white balls, total $9$ balls. After transfer, Bag II will have $10$ balls.

Case 1: Transferred ball is red

Then Bag II becomes:

• red $=3$,
• black $=5$,
• white $=2$.

Hence, $P(D_B\mid T_R)=\frac{5}{10}=\frac{1}{2}.$

Case 2: Transferred ball is black

Then Bag II becomes:

• red $=2$,
• black $=6$,
• white $=2$.

Hence, $P(D_B\mid T_B)=\frac{6}{10}=\frac{3}{5}.$

Case 3: Transferred ball is white

Then Bag II becomes:

• red $=2$,
• black $=5$,
• white $=3$.

Hence, $P(D_B\mid T_W)=\frac{5}{10}=\frac{1}{2}.$

3. Compute $P(D_B)$ using total probability

$P(D_B)=P(T_R)P(D_B\mid T_R)+P(T_B)P(D_B\mid T_B)+P(T_W)P(D_B\mid T_W).$

Substitute values: $P(D_B)=\frac{3}{10}\cdot\frac{1}{2}+\frac{2}{5}\cdot\frac{3}{5}+\frac{3}{10}\cdot\frac{1}{2}.$

$P(D_B)=\frac{3}{20}+\frac{6}{25}+\frac{3}{20}.$

$\frac{3}{20}+\frac{3}{20}=\frac{6}{20}=\frac{3}{10}.$

So, P(D_B)=\frac{3}{10}+\frac{6}{25}=rac{15}{50}+\frac{12}{50}=\frac{27}{50}.

4. Apply Bayes' theorem