Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability, that the ball drawn is white, is 29/45 , then n is equal to :

Bag 1: 4 white, 5 black (total 9). Bag 2: n white, 3 black (total n+3).

Case 1: White ball transferred from Bag 1 to Bag 2.

Probability = $$\frac{4}{9}$$. Bag 2 becomes: (n+1) white, 3 black (total n+4).

P(white from Bag 2) = $$\frac{n+1}{n+4}$$

Case 2: Black ball transferred from Bag 1 to Bag 2.

Probability = $$\frac{5}{9}$$. Bag 2 becomes: n white, 4 black (total n+4).

P(white from Bag 2) = $$\frac{n}{n+4}$$

Total probability of drawing white from Bag 2:

$$ P = \frac{4}{9} \cdot \frac{n+1}{n+4} + \frac{5}{9} \cdot \frac{n}{n+4} = \frac{4(n+1) + 5n}{9(n+4)} = \frac{9n+4}{9(n+4)} $$

Setting equal to $$\frac{29}{45}$$:

$$ \frac{9n+4}{9(n+4)} = \frac{29}{45} $$

$$ \frac{9n+4}{9(n+4)} = \frac{29}{45} $$

Cross-multiplying: $$45(9n+4) = 29 \times 9(n+4)$$

$$405n + 180 = 261n + 1044$$

$$144n = 864$$

$$n = 6$$

The correct answer is Option 1: 6.