Urn A contains seven white balls and five black balls. Urn B contains four white balls and six black balls. A ball is drawn from Urn A and then transferred to Urn B. A ball is then drawn from Urn B. What is the probability that the transferred ball was black given that the second ball drawn was black

P(X/Y) = 0.4545

Step-by-step explanation:

Let's call X that the transferred ball is black, X' that the transferend ball is white and Y the event that the second ball drawn is black.

The probability that the transferred ball was black given that the second ball drawn was black is equal to:

P(X/Y) = P(X∩Y)/P(Y)

Where P(Y) = P(X∩Y) + P(X'∩Y)

Then, the probability P(X∩Y) that the transferred ball is black and the second ball drawn is black is equal to:

P(X∩Y) = (5/12)*(7/11) = 0.2651

Because the urn A has 12 balls and 5 of them are black, then if the transferred ball is black, the urn B will contain 11 balls and 7 of them will be black balls.

At the same way, the probability P(X'∩Y) that the transferred ball is white and the second ball drawn is black is equal to:

P(X'∩Y) = (7/12)*(6/11) = 0.3181

Because the urn A has 12 balls and 7 of them are white, then if the transferred ball is white, the urn B will contain 11 balls and 6 of them will be black balls.

So, P(Y)  and P(X/Y) are equal to:

P(Y) = 0.2651 + 0.3181

P(Y) = 0.5832

P(X/Y) = 0.2651/0.5832

P(X/Y) = 0.4545

step-by-step explanation:

ok so im not 100% sure but you would start by adding 5.1t and 1.9t beacuse they have the same variable. that is 7t.   then add -3.6 + 1.2 which is -2.4

so then keep that as -2.4 + 7t. i apologize if this is incorrect. best luck on your test < 3