# Urn with white and black balls, random variable, conditional probability

An urn contains white and black balls with $p_w=p$ and $p_b=1−p$. If I made some extractions with replacement, what are the support and the probability function of $X_a$, where $X_a$ is the random variable representing the number of extraction made for get the a-th white ball? How can I get the conditional probability $P(X_a=s|X_b=t)$ (i.e. $X_9=21|X_5=11)$?

I have no idea of how to proceed...

• Assuming $a \gt b$ and $s \gt t$, and that $p$ is known, I would have guessed $P(X_a=s|X_b=t)=P(X_{a-b} = s-t)$, e.g. $X_4=10$ in your example, with a negative binomial distribution – Henry Feb 9 '16 at 17:33
• Probability should be between 0 and 1, so I think this is not correct. – Paul Feb 9 '16 at 17:45
• @Paul, Henry means $\Pr(X_9=21\mid X_5=11)=\Pr(X_4=10)$ where $X_k\sim\mathcal{NB}(k, p_w)$ – Graham Kemp Feb 10 '16 at 1:00

A start: The number of draws until we get the first white is at least $a$, and could be very large. Let us find $\Pr(X_a=k)$ for $k\ge a$.
We have $X_a=k$ if (i) we have exactly $a-1$ white on the first $a-1$ draws and (ii) we get a white on the $a$-th draw.
The probability of exactly $a$ white in the first $k-1$ draws is (binomial distribution) $\binom{k-1}{a-1}p^{a-1}(1-p)^{(k-1)-(a-1)}$. For $\Pr(X_a=k)$, multiply by $p$. We get $$\Pr(X_a=k)=\binom{k-1}{a-1}p^a(1-p)^{k-a}.$$ For more detail about the distribution of $X_a$, please search under negative binomial distribution.
Now perhaps you can tackle the condition prrobability problems on your own. You will not necessarily need this, but recall that $$\Pr(X_a=s\mid X_b=t)=\frac{\Pr(X_a=s\cap X_b=t)}{\Pr(X_b=t)}.$$ It will be worthwhile to tackle first a specific numerical example, like the one in the post.
• Hi, I've looked for a while about conditional negative binomial without any useful result and now I'm stuck on $P(X_a=s\cap X_b=t)$ – Paul Feb 14 '16 at 18:07