A bin has 8 black balls and 7 white balls. 3 of the balls are drawn at random. What is the probability of drawing 2 of one color and 1 of the other color?
Here's what I tried:
Case 1: 2 black balls and 1 white ball
8/15 * 7/14 * 7/13 = 392/2730 = 28/195
Case 2: 2 white balls, 1 black ball
7/15 * 6/14 * 8/13 = 336/2730 = 24/195
28/195 + 24/195 = 52/195 = 4/15
My teacher said it was wrong and I don't get why. She didn't give me the solutions either and just told me the answer was 4/5. Can someone tell me what I'm doing wrong? Thanks!
Best Answer
Case $1$: $2$ black balls and $1$ white ball. The correct probability will be
$P(2B,1W) = \displaystyle \frac{ 3 \cdot8 \cdot 7 \cdot 7}{15 \cdot 14 \cdot 13} = \frac{28}{65}$.
$\big[\text{It is basically } \frac{8}{15} \cdot \frac{7}{14} \cdot \frac{7}{13} + \frac{8}{15} \cdot \frac{7}{14} \cdot \frac{7}{13} + \frac{7}{15} \cdot \frac{8}{14} \cdot \frac{7}{13}\big]$
I would suggest think as selecting $2$ black balls out of $8$ and $1$ white ball out of $7$ vs. selecting any $3$ balls out of $15$ which can be written as -
$P(2B,1W) = \displaystyle {8 \choose 2}{7 \choose 1} / {15 \choose 3}$
Case $2$: $2$ white balls, $1$ black ball
$P(2W,1B) = \displaystyle {8 \choose 1}{7 \choose 2} / {15 \choose 3} = \frac{24}{65}$.
Adding them, we get probability as $\displaystyle \frac{4}{5}$.