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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):

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

Similarly:

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

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?

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    $\begingroup$ Absolutely correct, and well done. $\endgroup$ – André Nicolas Apr 5 '13 at 18:24
  • $\begingroup$ 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 $P(B)$. $\endgroup$ – Brian M. Scott Apr 5 '13 at 18:46
  • $\begingroup$ You are right, thanks! $\endgroup$ – mak Apr 5 '13 at 20:09
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    $\begingroup$ Fix the (minor) problems in your question, post it as an answer and accept it. $\endgroup$ – vonbrand Feb 8 '14 at 0:13
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Yes, that is correct.

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

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