one box contains 7 red and 4 green balls, and the second box contains 3 red and 5 green balls. one ball is drawn from the first box and placed unseen in the second box. what is the probability that a ball now drawn from the second box is red

We can divide the event that the second ball drawn is red into 2 disjoint events, and obtain its probability by summing both probabilities for the disjoint cases.

Case 1: The ball drawn from the first box is red and the ball drawn for the second box is also red
Case 2: The ball drawn from the first box is green but the ball drawn from the second box is red

For the first case, we have 11 balls in the first box, 7 of them being red. Thus, the probability of drawing a red ball is 7/11. After that, we will put the red ball drawn in the second box, so that it will have 9 balls in total, 4 of them being red. As a consecuence, the probability of drawing a red ball in this instance is 4/9. We need to multiply that probability by the probability that a red ball was drawn first, so we have a probability of 7/11 * 4/9 = 28/99 that the balls drawn are both red.

Now, lets go to the second case. The probability of drawing a green ball from the first box is 4/11, and after placing that ball in the second box, the probability that the new ball drawn is red is 3/9 = 1/3; this is because we will have 3 red balls out of a total of 9 (6 are green). Thus, the probability of this second event is 4/11 * 1/3 = 4/33.

Combining both probabilities, the probability that the ball drawn in the second box is red is 28/99 +4/33 = 40/99 = 0.40404040404...