Well the probability of an event is:
number of favorable cases
------------------------------
number of possible cases

For your first draw, you're looking to get a black ball. There are (x + y) balls to choose from, but only x are favorable (cause you have only x black balls), with probability x/(x + y).

For your second draw, you have (x + y - 1) balls left, since you've already extracted one ball. You're trying to get a white ball, and there are y white balls left if you consider that your first draw was successful. In this case, the probability will be y/(x + y - 1).

The probability of both events will be obtained by multiplying the two probabilities above.

uptowngirl92 wrote:A jar contains only x black balls and y white balls. One ball is drawn randomly from the jar and is not replaced. A second ball is then drawn randomly from the jar. What is the probability that the first ball drawn is black and the second ball drawn is white?

(A) [x/(x+y)] [y/(x+y)]

(B)[x/(x+y)] [(x-1)/(x+y-1)]

(C) xy/x+y

(D) [(x-1)/(x+y)] [(y-1)/(x+y)]

(E)[x/(x+y)] [y/(x+y-1)]

The total number of balls in the jar is (x + y). The probability of drawing a black ball followed by a white ball, without replacement, is:

x/(x + y) * y/(x + y - 1)

Answer: E