# Probability, choose a box and then take exactly two white balls

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?

• 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