# [Math] Probability of drawing a white ball out of Box A

probability

I have been looking through some course notes to review some probability theory, as I have not taken a statistics course in a couple of semesters. I came across this problem:

In box A there are two red balls and one white ball. In box B there
are three red balls and two white balls. If you randomly chose one box
and randomly draw a ball out of it, and the ball is white, what is the
probability that you chose box A?

The given solution is:

The probability that you draw a white ball out of box A is
$\frac{1}{3}$ Similarly, the probability that you draw a white ball
out of box B is $\frac{2}{5}$.

If you randomly choose a box, then the probability you draw a white
ball is:

$\frac{1}{2}\cdot\frac{1}{3} + \frac{1}{2}\cdot\frac{2}{5} = \frac{11}{30}$

Since, of this probability, the probability that you draw a white ball
out of box A is $\frac{1}{2}\cdot\frac{1}{3} = \frac{1}{6}$ the
probability that you chose box A once you drew a white ball is:

$\frac{1}{6}/\frac{11}{30} = \frac{5}{11}$

My question is, why is the probability of drawing a white ball $\frac{11}{30}$ (and not $\frac{3}{8}$)? I understand the remainder of the solution when I accept the probability of drawing a white ball as $\frac{11}{30}$, I just don't understand why this is the case.

You showed the calculation. It came out to $\frac{11}{30}.$ That is why the probability is $\frac{11}{30}.$
The reason you cannot just divide the number of white balls by the total number of balls is that not every ball has an equal chance of being selected. Each ball in box $B$ is less likely to be selected than each ball in box $A$. This must be the case, because "I selected one of the balls that came from box $A$" has probability $\frac12,$ whereas if the balls all had equal chance to be picked, box $A$ would have only $\frac38$ probability.