Final answer:
The probability that the transferred balls include 1 black is 3/7.
Step-by-step explanation:
To calculate the probability that the transferred balls and 1 black, we need to consider the possible outcomes.
Case 1: The transferred balls are both black. There are 3 black balls in bag A and 2 black balls in bag B, so the probability is
(3/7) * (2/4) = 6/28.
Case 2: One transferred ball is black and the other is white. There are 3 black balls in bag A and 2 white balls in bag B, so the probability is
(3/7) * (2/4) = 6/28.
Adding the probabilities from Case 1 and Case 2, the probability that the transferred balls include 1 black is
6/28 + 6/28 = 12/28 = 3/7.