Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There are $5$ boxes. There are $5$ white and $3$ black balls in two boxes, and $4$ white and $6$ black balls in the other three boxes. One box is randomly chosen. $3$ balls are randomly taken from the chosen box.

What is the probability that exactly $2$ of the chosen balls are white?

  • $A$ - the box with $8$ balls is chosen
  • $\bar{A}$ - the box with $10$ balls is chosen
  • $B$ - exactly two chosen balls are white

There are $5$ boxes, $2$ boxes with $8$ balls: $2/5$. Choosing the box and taking the balls are independent events so I can multiply the probabilities. There are $8$ balls in the box, I need to take $3$ balls $\binom83$, of which $2$ are white $\binom52$ and $1$ black $\binom31$ (there are $5$ white balls and $3$ black balls):

$$P(B \mid A)=\frac{2}{5} \cdot \frac{\dbinom52 \dbinom31}{\dbinom83}$$


$$P(B \mid \bar{A})=\frac{3}{5} \cdot \frac{\dbinom42 \dbinom61}{\dbinom{10}3}$$

So now I can calculate $P(B)$:

$$P(B)=P(B \mid A) \cdot P(A)+P(B \mid \bar{A}) \cdot P(\bar{A})$$

Is this correct?

share|cite|improve this question
Absolutely correct, and well done. – André Nicolas Apr 5 '13 at 18:24
There is a small error in what you’ve written, though probably not in what you were thinking: $P(B\mid A)$ is just $$\frac{\binom52\binom31}{\binom83}\;,$$ without the $\frac52$. The $\frac25$ is the $P(A)$ that you want when you calculate $P(B)$. – Brian M. Scott Apr 5 '13 at 18:46
You are right, thanks! – mak Apr 5 '13 at 20:09
Fix the (minor) problems in your question, post it as an answer and accept it. – vonbrand Feb 8 '14 at 0:13
up vote 0 down vote accepted

Yes, that is correct.

(Really, I just wanted this not to show up as "unanswered" on the (probability) tag's first page.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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