# Calculating individual probabilities in marginalization and bayes' rule

I have a problem and a solution to it that I don't quite understand.

Question: Five identical bowls are labeled 1, 2, 3, 4, and 5. Bowl i contains i white and 5 − i black balls, with i = 1, 2, . . . , 5. A bowl is randomly selected and two balls are randomly selected (without replacement) from the contents of the bowl.

a. What is the probability that both balls selected are white?

b. Given that both balls selected are white, what is the probability that bowl 3 was selected?

Solution: Let $A = \{both\,balls\,are\,white\}$ and for $i=1,2,3,4,5$

$A_i$ = both balls selected from bowl $i$ are white, then $\bigcup A_i = A$

$B_i$ = bowl $i$ is selected. Then $P(B_i) = .2$ for all $i.$

a. $P(A) = \sum P(A_i|B_i)P(B_i) = \frac15[0+\frac25(\frac14)+\frac35(\frac24)+\frac45(\frac34)+1]$ (I understand the sum equation (marginalization), but how are the individual probabilities calculated?)

b. Using Bayes' rule, $P(B_3|A)=\frac{\frac3{50}}{\frac2{50}}$ (How are the individual probabilities calculated?)

Any pointers to the correct approach are appreciated!

Ok, I figured it out. For P(A), the outer 1/5 is the $P(B_i)$ and the inner five elements are $P(A_i|B_i)$ for each i. Here, $P(A_1|B_1)$ is zero cuz the probability of selecting the second white ball is zero. $P(A_2|B_2)$ is $\frac25(\frac14)$ cuz the probability of taking out the first white ball is $\frac25$ and the second one is $\frac14$ and so on. About $P(B_3|A)$, there seems to be a printing mistake. The correct value is $\frac{P(B_3 \cap A)}{P(A)}$, i.e. $\frac{\frac3{50}}{\frac2{5}}$.