3 balls in a box We have a box with $3$ balls, that can be black or white. 
We extract a ball, and it's white. Then we put the ball in the box, we extract again a ball and it's white.
What is the probability that in the box there are $3$ white balls ? 
I obtain $\frac{9}{14}$  but I'm not sure about this result. 

Assume each ball has the same probability of being black or white. The prior probability of having $n$ white balls is $\binom 3n\frac18$, $n=0,1,2,3$.
Given there are $n$ white ball(s) in the box, then the probability of drawing two white balls with replacement is $\left(\frac n3\right)^2$.
The posterior probability of having 3 white balls, given the result, is:
$$\begin{align*}
P(W=3\mid W_1,W_2) &= \frac{P(W_1,W_2\mid W=3)\times P(W=3)}{P(W_1,W_2)}\\
 &= \frac{P(W_1,W_2\mid W=3)\times P(W=3)}{\sum_{n=0}^3P(W_1,W_2\mid W=n)\times P(W=n)}\\
 &= \frac{\left(\frac 33\right)^2\binom 33\frac18}{\sum_{n=0}^3\left(\frac n3\right)^2\binom 3n\frac18}\\
 &= \frac{3^2\times1}{0^2\times1+1^2\times3+2^2\times3+3^2\times1}\\
 &= \cdots
\end{align*}$$
