A bag I contains 4 white balls and 6 black balls. Another bag II contains 4 white balls and 3 black balls. One ball is selected at random from one of the bags and it is found to be black. Find the probability that it was drawn from bag I?

To solve this problem, we will use Bayes' theorem, which is a way to find a probability when we know certain other probabilities. The formula is:

P(A|B) = [P(B|A) * P(A)] / P(B)

Where:

• P(A|B) is the probability of event A given event B is true
• P(B|A) is the probability of event B given event A is true
• P(A) and P(B) are the probabilities of events A and B

Let's define:

• Event A1: The ball is drawn from Bag I
• Event A2: The ball is drawn from Bag II
• Event B: The ball drawn is black

We are asked to find P(A1|B), the probability that the ball was drawn from Bag I given that it is black.

From the problem, we know that:

• P(A1) = P(A2) = 1/2 (since one bag is selected at random)
• P(B|A1) = 6/10 (since there are 6 black balls out of a total of 10 balls in Bag I)
• P(B|A2) = 3/7 (since there are 3 black balls out of a total of 7 balls in Bag II)

We also know that P(B) can be found using the law of total probability: P(B) = P(B and A1) + P(B and A2) = P(B|A1)P(A1) + P(B|A2)P(A2) = (6/10 * 1/2) + (3/7 * 1/2) = 1/2

Now we can substitute these values into Bayes' theorem to find P(A1|B):

P(A1|B) = [P(B|A1) * P(A1)] / P(B) = [(6/10) * (1/2)] / (1/2) = 6/10 = 0.6

So, the probability that the black ball was drawn from Bag I is 0.6 or 60%.