An honest die is thrown 8 times; let $X$ be the number of twos and let $Y$ be the number of fours. Find the joint pmf of $X$ and $Y$ and calculate $\mathbb{P}(X=Y)$.
This question is too big to manually draw a table for so I'm having trouble solving it. Can anyone help with the joint pmf? Calculating $\mathbb{P}(X=Y)$ should be straightforward once I have the joint pmf.
I know that for either $X$ and $Y$ alone, the pmf will be ${{8}\choose{k}}(1/6)^k(5/6)^{8-k}$
The joint distribution is not just $\Pr[X=x,Y=y]=\Pr[X=x| Y=y]\Pr[Y=y]$?
The last form is easy to handle, right? You just need to be aware that $\Pr[X=x|Y=y]=\Pr[Z=z]$ for the random variable $Z$ that counts the number of two in $8-y$ die. However you need to take in account that these $8-y$ die must be all different of four.
That is: $\Pr[X=x|Y=y]$ means "probability that will be exactly $x$ two's in $8-y$ die, and the other $8-y-x$ die will be distinct of two or four".
Take a look here also. For a "shortcut" to an easier way to handle this joint distribution take a look at the multinomial distribution.