Bag I contains 3 red and 4 black balls, and Bag II contains 5 red and 2 black balls. Two balls are transferred at random from Bag I to Bag II and then a ball is drawn at random from Bag II. Find the probability that the drawn ball is red in colour.

$\frac{41}{63}$

The correct answer is Option (2) → $\frac{41}{63}$ ##

Let $E_1, E_2, E_3$ and $A$ be events:

$E_1$: Both transferred balls from Bag I to Bag II are red.

$E_2$: Both transferred balls from Bag I to Bag II are black.

$E_3$: Out of two transferred balls, one is red and other is black.

$A$: Drawn ball is red in colour.

Now, $P(E_1) = \frac{{}^3C_2}{{}^7C_2} = \frac{1}{7}$

$P(E_2) = \frac{{}^4C_2}{{}^7C_2} = \frac{2}{7}$

$P(E_3) = \frac{{}^3C_1 \times {}^4C_1}{{}^7C_2} = \frac{4}{7}$

$P\left(\frac{A}{E_1}\right) = \frac{7}{9}, P\left(\frac{A}{E_2}\right) = \frac{5}{9}$,

$P\left(\frac{A}{E_3}\right) = \frac{6}{9}$

Using Total Probability Theorem, we get

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

$= \frac{1}{7} \cdot \frac{7}{9} + \frac{2}{7} \cdot \frac{5}{9} + \frac{4}{7} \cdot \frac{6}{9} = \frac{7+10+24}{63} = \frac{41}{63}$