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

P=\frac {4}{10} \,\times\, \frac {3}{9} = 0.133.

The probabilty of drawing two blacks in a row is

P=\frac {6}{10} \,\times\, \frac {5}{9} = 0.333

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

0.133 + 0.333 = 0.467

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

P = 1 - 0.467 = 0.533

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:


1 - \frac 4 {10} \times \frac 3 9 - \frac 6 {10} \times \frac 5 9

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.

2\cdot \frac{4}{10} \cdot\frac{6}{9}=\not{2}^{1}\cdot\frac{4}{\not{10} _{5}}\cdot \frac{\not{6}^{2}}{\not{9}_{3}}=\frac{8}{15}