Question

An urn contains one white chip and a second chip that is equally likely to be white or black. A chip is drawn at random and returned to the urn. Then a second chip is drawn.What is the probability that a white appears on the second draw given that a white appeared on the first draw? (Hint: Let Wi be the event that a white chip is selected on the ith draw, i = 1, 2. Then P(W2|W1) = P(W1∩W2) P(W1) . If both chips in the urn are white, P(W1) = 1; otherwise, P(W1) = 1/2)

Answer 1

To find the probability of a white chip appearing on the second draw given that a white appeared on the first draw, we use the formula P(W2|W1) = P(W1∩W2) / P(W1). By calculating the values, we find that the probability is 1/4.

Step-by-step explanation:

To find the probability that a white appears on the second draw given that a white appeared on the first draw, we can use the formula P(W2|W1) = P(W1∩W2) / P(W1). Let's calculate:

P(W1∩W2) = (1/2) * (1/2) = 1/4

P(W1) = 1

P(W2|W1) = P(W1∩W2) / P(W1) = (1/4) / 1 = 1/4

Therefore, the probability that a white appears on the second draw given that a white appeared on the first draw is 1/4.

Answer 2

`P(W_(2) | W_(1))` = 1 if both chips are white,
`P(W_(2) | W_(1))` = 1/2 otherwise

Step-by-step explanation:

`W_(i)` represents the event that a white chip is selected on the ith draw. Then
`P(W_(2) | W_(1)) = P(W_(2)\cap W_(1))/P(W_(1))`. If both chips in the urn are white,
`P(W_(2)\cap W_(1)) = 1` and
`P(W_(1)) = 1`, then
`P(W_(2) | W_(1)) = P(W_(2)\cap W_(1))/P(W_(1)) = 1`. If one chip is white and the other is black, then
`P(W_(2) | W_(1)) = P(W_(2)\cap W_(1))/P(W_(1)) = P(W_(2))P(W_(1))/P(W_(1)) = P(W_(2)) = 1/2` because the event a white chip is selected on the first draw is independent from the event a white chip is selected on the second draw, and because the chips are drawn at random.