Conditional and Joint Probability

Question

Box 1 : 5 white and 2 black balls Box 2 : 2 white and 1 black balls Box 3 : 2 white and 3 black balls

One box is selected at random and one ball is drawn from it. What is the probability that it will be white?

I think the answer is 1/3(5/7 + 2/3 + 2/5) = 187/315.

Is this correct?

Also calculate the probability the first box was selected given that a white ball was drawn? I am stumped on this second questions. I think it is P(1 | W) which comes out to more than 1 which is obviously not right.

Any ideas?

Thanks

Answer 1

$\newcommand{\P}{\operatorname{\sf P}}$You've got the first part right. I'm not sure what you're doing on the second. You need to apply Bayes' Rule.

Let $W$ be the event of choosing a white ball, and $B$ be the number of the box chosen.

$$\begin{align} \text{Then by} & \text{ the law of total probability:} \\[1ex]\P(W) & = \P(B=1)\P(W\mid B=1)+\P(B=2)\P(W\mid B=2)+\P(B=3)\P(W\mid B=3) \\[1ex] & = \frac 1 3\Bigl(\P(W\mid B=1)+\P(W\mid B=2)+\P(W\mid B=3)\Bigr) \\[1ex] & = \frac 1 3\Bigl(\frac 5 7+\frac 2 3+ \frac 2 5\Bigr) \\[4ex] \text{Then apply} & \text{ Bayes' Rule:} \\[1ex] \P(B=1\mid W) & = \frac{P(B=1)P(W\mid B=1)}{P(W)} \\[1ex] & = \frac{\frac 1 3\cdot\frac 5 7}{\frac 1 3\Bigl(\frac 5 7+\frac 2 3+ \frac 2 5\Bigr)} \\[1ex] & = \frac{75}{75 + 70 + 42} \end{align}$$

Answer 2

The answer to your second question is indeed $P(1|W)=75/187$ which is less then $1$.
Using Bayes theorem,
$P(1|W)=\frac{P(W|1).P(1)}{P(W|1).P(1)+P(W|2).P(2)+P(W|3).P(3)}$
$P(1|W)=\frac{(5/7).(1/3)}{(5/7).(1/3)+(2/3).(1/3)+(2/5).(1/3)}$
$P(1|W)=75/187$
Hope this helps