4
$\begingroup$

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!

$\endgroup$
1
  • 2
    $\begingroup$ You need to multiply by $3$ as for example, in the first case, the white ball can be either the first, second or third. $\endgroup$ – Math Lover Feb 28 at 18:50
5
$\begingroup$

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 $\displaystyle \frac{4}{5}$.

$\endgroup$
4
$\begingroup$

What you have written corresponds to the case when first two balls of one color and then the ball of the other color are drawn. Therefore your calculations miss the factor 3 accounting for possible permutations of the balls.

$\endgroup$
4
$\begingroup$

As stated in the comments to your question, you have ascertained the probability of drawing two black balls and a white ball in that order and the possibility of drawing two white balls and a black ball in that order. You have to multiply by $3$ to account for all possible orders of drawing the balls you want.

Here's another approach. The only other possibility is to draw three balls of the same color. The probability of drawing three black balls is $\frac{8}{15} \cdot \frac {7}{14} \cdot \frac{6}{13}=\frac{336}{15 \cdot 14 \cdot 13}$ and the probability of drawing three white balls is $\frac{7}{15} \cdot \frac{6}{14} \cdot \frac{5}{13}=\frac{210}{15 \cdot 14 \cdot 13}$, so the probability of one of these two occurrences is $\frac{446}{15 \cdot 14 \cdot 13}=\frac 15$, and the probability that doesn't happen (i.e., you end up with two balls of one color and one ball of another color) must therefore be $1-\frac 15=\frac 45$.

$\endgroup$
1
$\begingroup$

The answer follows from the story of the Hypergeometric distribution! A random variable that counts the number of successes (in this case, let's call a success drawing a white ball) from a fixed number of draws (3, in our case) without replacement is said to have a Hypergeometric distribution. (If the scenario were instead with replacement, we'd have a Binomially distributed random variable on our hands.)

Let's let $X$ be this particular random variable. We know that at a minimum $X = 0$, and at a maximum, $X = 3$, but we're interested in the probability that $X = 2$ (2 white balls, 1 black ball) or $X = 1$ (1 white ball, 2 black balls). Since these cases are disjoint, we can simply add the probabilities of these two cases: $P(X=1) + P(X=2)$.

There are $\binom{15}{3}$ ways to pick 3 balls. To get exactly 2 white balls (and therefore, 1 black ball), we need to choose 2 white balls from the 7 possible white balls, and 1 black ball from the 8 possible black balls. The same logic can be used to evaluate the other case:

$$ \begin{align} P(X=1) &= \frac{\binom{7}{1}\binom{8}{2}}{\binom{15}{3}}\\ P(X=2) &= \frac{\binom{7}{2}\binom{8}{1}}{\binom{15}{3}} \end{align} $$

Adding the probabilities together, we arrive at:

$$ P(X=1 \cup X=2) = \frac{4}{5} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.