A bowl contains six black balls and four white balls. Two balls are drawn at random one at a time without replacement. Find the probability of drawing a white and a black ball.:confused:

Ten balls in the bowl. Six are black; four are white. The first time you draw, the probability of drawing a black ball is 6/10; the probability of drawing a white ball is 4/10.

You want the probability of drawing both a white and a black ball. It doesn't matter which order they're drawn. The probability of drawing two whites in a row is

.

The probabilty of drawing two blacks in a row is



The probability of drawing either two blacks or two whites (two of the same color) is



The probability, therefore, of drawing one ball of each color is

Perito - your answer is incorrect, as you neglected that the two balls are selected without replacement. The probability of drawing two whites in a row is 4/10 * 3/9, and the probability of drawing two blacks in a row is 6/10 * 5/9. Hence the probability of drawing one white and one black is:

Thanks. I was thinking of it before I started typing, but it skipped my mind after that.

Here's another small variation.

The white probability is 4/10

Then, the black would be 6/9

Since there are 2 different ways they can be chosen... black-white or white-black, we multiply by 2.