A pot contains 5 red and 2 green balls. A ball is drawn at random from this pot. If a drawn ball is green, then a red ball is added to the pot. If a drawn ball is red, then a green ball is added to the pot, while the original ball drawn is not replaced in the pot. Now a second ball is drawn at random from the pot, what is the probability that the second ball drawn is a red ball?

Step-by-step solution

Let R₁ and G₁ be the events that the first ball drawn is red and green, respectively. Initially, the pot contains 5 red and 2 green balls, total 7 balls. Probability of drawing a red ball first: P(R₁) = 5 7 . Probability of drawing a green ball first: P(G₁) = 2 7 . Case 1: If the first ball is red ( R₁ ), it is not replaced, and a green ball is added. The pot now contains (5-1) red and (2+1) green balls, i.e., 4 red and 3 green balls. Total balls = 7. The probability of drawing a red ball second given R₁ is P(R₂ |