What is the probability of no white balls There are $n$ balls in a box. We add to the box a white ball. After that the ball drawn from the box turns out to be white. What is the probability that initially there was no white balls in the box?
I try to apply the Bayes formula:
$$\mathbb{P}(A|B) = \frac{\mathbb{P}(B|A) \mathbb{P}(A)}{\mathbb{P}(B)}$$
$A = \text{initially there was no white balls}$
$B = \text{the drawn ball after adding 1 white ball to the box is white}$
$\mathbb{P}(B|A) = \text{the probability that the drawn ball is white if initially there was no white balls} = \frac{1}{n+1}$
Am I right that $\mathbb{P}(A) = \frac{1}{n+1}$ and $\mathbb{P}(B) = \frac{k+1}{n+1}$, where $k$ denotes the initial numer of white balls?
Thanks in advance.
 A: This question cannot be answered without an assumption about your prior knowledge of the balls in the box. I'll work it out for two relatively plausible assumptions to show what a difference the assumption can make.
If your prior beliefs were indifferent with respect to the number of white balls in the box, you'd assign a prior probability of $\frac1{N+1}$ to each of the $N+1$ possible numbers of white balls. Then the posterior probability for there to have been no white balls before you added one is
\begin{eqnarray*}
\mathsf P(N=0\mid\text{white ball drawn})
&=&
\frac{\mathsf P(N=0\land\text{white ball drawn})}{\mathsf P(\text{white ball drawn})}
\\
&=&
\frac{\frac1{N+1}\cdot\frac1{N+1}}{\sum_{n=0}^N\frac1{N+1}\cdot\frac{n+1}{N+1}}
\\
&=&
\frac2{(N+1)(N+2)}\;,
\end{eqnarray*}
so it decays quadratically with $N$.
On the other hand, if you drew the balls that were already in the box from a huge box filled equally with white and non-white balls, so for each ball independently you're indifferent whether it's white or not, we get
\begin{eqnarray*}
\mathsf P(N=0\mid\text{white ball drawn})
&=&
\frac{\mathsf P(N=0\land\text{white ball drawn})}{\mathsf P(\text{white ball drawn})}
\\
&=&
\frac{2^{-N}\cdot\frac1{N+1}}{\sum_{n=0}^N2^{-N}\binom Nn\cdot\frac{n+1}{N+1}}
\\
&=&
\frac{2^{-(N-1)}}{N+2}\;,
\end{eqnarray*}
so the posterior probability decays exponentially with $N$.
