A bag contains 4 red and 6 green balls. A ball is drawn at random, its colour is noted and is returned to the bag. One additional ball of the colour drawn is put in the bag. Again a ball is then drawn from the bag. The probability of this ball to be of green colour is

$\frac{3}{5}$

The correct answer is Option (3) → $\frac{3}{5}$

Initial composition: 4 red and 6 green balls.

Total = 10 balls.

Let $G_1$ be event that first draw is green, $R_1$ be event that first draw is red.

Case 1: First draw is green.

Probability = $\frac{6}{10}$

New composition: 4 red, 7 green → total = 11 balls

Probability of green on second draw = $\frac{7}{11}$

Case 2: First draw is red.

Probability = $\frac{4}{10}$

New composition: 5 red, 6 green → total = 11 balls

Probability of green on second draw = $\frac{6}{11}$

Total probability of second ball being green:

$P(G_2) = \frac{6}{10} \cdot \frac{7}{11} + \frac{4}{10} \cdot \frac{6}{11} = \frac{42 + 24}{110} = \frac{66}{110} = \frac{3}{5}$