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?
Best Answer
Yes, that is correct.
(Really, I just wanted this not to show up as "unanswered" on the (probability) tag's first page.)