Probability: In n rolls of a fair die

Question

...Let X be the number of times 1 is rolled and Y be the number of times a 2 is rolled. Find the conditional distribution of X given Y=y.

My work thus far:

1. $$ P(X=x | Y=y) = \frac{P(X=x , Y=y)}{P(Y=y)} $$

2. $$P(Y=y) = \begin{pmatrix} n\\ y\\ \end{pmatrix}\frac{5}{6}^{n-y}\frac{1}{6}^y$$

Am I going about this correctly? How do I calculate the joint distribution? Would it be correct to have it as:

$$P(X=x , Y=y) = \begin{pmatrix} n\\ y\\ \end{pmatrix} (\frac{1}{6})^x (\frac{1}{6})^y$$

This seems wrong to me though.. Thank you for your help in advance.

Answer 1

You don't need the joint distribution because given the condition the conditional probability can be calculated.

If $Y=y$ (the number of the twos) is given then there are $n-y$ positions for the $x$ pcs of ones. That is, the conditional probability in question is

$$P(X=x\mid Y=y)={n-y\choose x}\left(\frac16\right)^x\left(\frac56\right)^{n-y-x}.$$ ($x=0,1,\dots,n-y$ and $y=0,1,\dots, n.$)

This is a simple binomial distribution with $n-y$ trials and $p=\frac16$ as success probability.

Answer 2

The zoli answer is almost right. But I think you have to consider that the number $2$ wont't be rolled in the remaining $n - y$ rolls. This way, the probability of the number $1$ is rolled will be $1/5$. So, the final result is $$P(X=x\mid Y=y)={n-y\choose x}\left(\frac15\right)^x\left(\frac45\right)^{n-y-x}.$$ ($x=0,1,\dots,n-y$ and $y=0,1,\dots, n.$)