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

The bag initially contains 4 red balls and 6 green balls, for a total of 10 balls.

A ball is drawn, its color is noted, it is returned to the bag, and one additional ball of the same color is added. Then a second ball is drawn.

The second ball can be green in two scenarios: either the first ball drawn was red, or the first ball drawn was green.

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If the first ball drawn is red:

Probability of drawing a red ball first:

$P(\text{1st ball is Red}) = \frac{4}{10}$

$P(\text{1st ball is Red}) = \frac{2}{5}$

After returning the red ball and adding one more red ball, the bag contains 5 red balls and 6 green balls, for a total of 11 balls.

Probability of drawing a green ball second:

$P(\text{2nd ball is Green | 1st was Red}) = \frac{6}{11}$

Combined probability for this scenario:

$P(\text{Red then Green}) = \frac{2}{5} \times \frac{6}{11}$

$P(\text{Red then Green}) = \frac{12}{55}$

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If the first ball drawn is green:

Probability of drawing a green ball first:

$P(\text{1st ball is Green}) = \frac{6}{10}$

$P(\text{1st ball is Green}) = \frac{3}{5}$

After returning the green ball and adding one more green ball, the bag contains 4 red balls and 7 green balls, for a total of 11 balls.

Probability of drawing a green ball second:

$P(\text{2nd ball is Green | 1st was Green}) = \frac{7}{11}$

Combined probability for this scenario:

$P(\text{Green then Green}) = \frac{3}{5} \times \frac{7}{11}$

$P(\text{Green then Green}) = \frac{21}{55}$

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The total probability that the second ball is green:

$P(\text{2nd ball is Green}) = \frac{12}{55} + \frac{21}{55}$

$P(\text{2nd ball is Green}) = \frac{33}{55}$

$P(\text{2nd ball is Green}) = \frac{3}{5}$

Therefore, the probability of the second ball being green is $\frac{3}{5}$.