Colored balls in three boxes (conditional probability problem) Problem
Suppose there are three boxes numbered with twenty balls in each of them. The first box contains twenty white balls; the second, fifteen, and the third,ten; the rest of the balls are black. One box is chosen randomly and two balls are extracted from it with replacement. The probability of choosing the first box is the same as the probability of choosing the second, where as the probability of selecting the third box is equal to the sum of the other two.
Calculate the probability of having chosen the first box knowing that the two extracted balls are white.
The attempt at a solution
I think this problem is related to conditional probabilities. Let $A_i=\{\text{the event of having chosen the i-th box}\}$. $B=\{\text{the two extracted balls are white}\}$. Then, we want to calculate $P(A_1|B)$. But $$P(A_1|B)=\frac{P(A_1 \cap B)}{P(B)}.$$
Note that since the first box contains only white balls, then $P(A_1 \cap B)=P(A_1)$. Since $\Omega=A_1 \amalg A_2 \amalg A_3$, then $1=P(\Omega)=P(A_1)+P(A_2)+P(A_3)=4P(A_1)$. From here we get $P(A_1)=\frac{1}{4}.$
Now, $B=B_1 \amalg B_2 \amalg B_3$, where $B_i=\{\text{select two white balls from the i-th box}\}$. So $P(B)=P(B_1)+P(B_2)+P(B_3)=\frac{1}{4}+\frac{1}{4}\frac{15}{20}\frac{15}{20}+\frac{1}{2}\frac{10}{20}\frac{10}{20}.$
Joining these two results, we have that $$P(A_1|B)=\dfrac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{4}\frac{15}{20}\frac{15}{20}+\frac{1}{2}\frac{10}{20}\frac{10}{20}}.$$
I am just learning this subject and I feel extremely insecure about my answers, that is why I would truly appreciate if someone could tell me if my solution is correct.
 A: $\color{green}\checkmark $ Your solution is fine.  Here's mine.
$\begin{align}
\text{Given:}\quad&\text{ Selection with replacement, 2 balls}
\\
 {\sf P}(B\mid A_1) & = (\frac{20}{20})^2 
\\[1ex]
 {\sf P}(B\mid A_2) & = (\frac{15}{20})^2 
\\[1ex]
 {\sf P}(B\mid A_3) & = (\frac{10}{20})^2 
\\[1ex]
 {\sf P}(A_1)& ={\sf P}(A_2)
\\
 {\sf P}(A_3)& = {\sf P}(A_1) + {\sf P}(A_2)\\[3ex]
\text{Find:}\quad & \text{the probability of: $A_1$ knowing that $B$}
\\
{\sf P}(A_1\mid B) &= \frac{{\sf P}(A_1\cap B)}{{\sf P}(B)}
\\[1ex] & = \frac{{\sf P}(A_1\cap B)}{{\sf P}(A_1\cap B)+{\sf P}(A_2\cap B)+{\sf P}(A_3\cap B)}
\\[1ex] & = \frac{{\sf P}(A_1){\sf P}(B\mid A_1)}{{\sf P}(A_1){\sf P}(B\mid A_1)+{\sf P}(A_2){\sf P}(B\mid A_2)+{\sf P}(A_3){\sf P}(B\mid A_3)}
\\[1ex] & = \frac{{\sf P}(B\mid A_1)}{{\sf P}(B\mid A_1)+{\sf P}(B\mid A_2)+2\,{\sf P}(B\mid A_3)}
\\[1ex] & = \frac{(\frac{20}{20})^2}{(\frac{20}{20})^2+(\frac{15}{20})^2+2(\frac{10}{20})^2}
\\[1ex] & = \frac{4^2}{4^2+3^2+2^3}
\\[1ex] & = \frac{16}{33}
\end{align}$
