Question

There are two bags, Bag I and Bag II. Bag I contains 3 red and 5 black balls, while Bag II contains 4 red and 3 black balls. One ball is transferred randomly from Bag I to Bag II, and then a ball is drawn randomly from Bag II. If the ball drawn is black, what is the probability that the transferred ball from Bag I is also black?

Answer 1

The probability that the transferred ball from Bag I is black given that the drawn ball from Bag II is black is 5/13.

Step-by-step explanation:

To find the probability that the transferred ball from Bag I is black given that the drawn ball from Bag II is black, we need to use conditional probability. Let's define the following events: R1 = ball transferred from Bag I is red, B2 = ball drawn from Bag II is black.

1. To find the probability of B2, we can use the law of total probability: P(B2) = P(B2 given R1) * P(R1) + P(B2 given not R1) * P(not R1). In this case, P(B2 given R1) = (3+1)/(8+1) = 4/9, P(R1) = 5/12, P(B2 given not R1) = 3/10, and P(not R1) = 7/12. Plugging in these values, we get P(B2) = (4/9) * (5/12) + (3/10) * (7/12) = 13/30.
2. To find the probability of both B2 and transferred ball being black, we multiply the probability of B2 given transferred ball being black by the probability of transferred ball being black: (4/9) * (3/8) = 1/6.
3. Finally, we can use conditional probability formula to find the probability of transferred ball being black given that B2 is black: P(black transferred ball given B2) = P(both black) / P(B2) = (1/6) / (13/30) = 5/13.