A coin is tossed. if it turns up head 2 balls will be drawn from urn A otherwise 2 balls will be drawn from a urn B. Urn A contains 3 black and 5 white balls. Urn B contains 7 black balls and 1 white ball. In both cases selections are to be made with replacement. what is the probability that Urn A is used given that both the balls drawn are black. The correct answer is 9/58?

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Problem: A coin is tossed. If it turns up head, 2 balls will be drawn from urn A otherwise 2 balls will be drawn from urn B. Urn A contains 3 black and 5 white balls. Urn B contains 7 black balls and 1 white ball. In both cases selections are to be made with replacement. What is the probability that Urn A is used given that both the balls drawn are black?

Solution:
Let's define the events:
- H: the coin turns up head
- T: the coin turns up tail
- A: 2 balls are drawn from urn A
- B: 2 balls are drawn from urn B
- BB: both balls drawn are black

We need to find P(A|BB), the probability that urn A is used given that both the balls drawn are black.

We can use Bayes' theorem to solve this problem:
P(A|BB) = P(BB|A) * P(A) / P(BB)

We know that P(A) = P(H) = 1/2, as the coin is fair. We also know that P(BB|A) is the probability of drawing 2 black balls from urn A, with replacement:
P(BB|A) = (3/8)^2 = 9/64

To calculate P(BB), we need to use the law of total probability:
P(BB) = P(BB|A) * P(A) + P(BB|B) * P(B)

We know that if the coin turns up tail, we draw 2 balls from urn B, so:
P(BB|B) = (7/8)^2 = 49/64
P(B) = P(T) = 1/2

Therefore:
P(BB) = (9/64) * (1/2) + (49/64) * (1/2) = 29/64

Now we can substitute all the values in Bayes' theorem:
P(A|BB) = (9/64) * (1/2) / (29/64) = 9/29

Therefore, the probability that urn A is used given that both the balls drawn are black is 9/29.

Answer

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