<s>[INST] Probability of two random variables being equal The question is as follows:

Let $X_1$ ~ Binomial(3,1/3) and $X_2$ ~ Binomial(4,1/2) be independent random variables. Compute P($X_1$ = $X_2$)

I'm not sure what it means to compute the probability of two random variables being equal.
 [/INST] Let $Z=X_1-X_2$
$P(Z=z)=\sum_{x_1=z+x_2}^\infty P(X_1=x,X_2=x-z)$ (since $X_1$ and $X_2$ are independent)
$P(Z=z)=\sum_{x_1=z+x_2}^\infty P(X_1=x)P(X_2=x-z)\\=\sum_{x_1=z+x_2}^\infty \binom{3}{x}(\frac{1}{3})^x(\frac{2}{3})^{3-x}\binom{4}{4-x+z}(\frac{1}{2})^{x-z}(\frac{1}{2})^{4-x+z}$
When $X_1=X_2\Rightarrow z=0 $
Therefore
$P(Z=0)=\sum_{x_1=x_2}^\infty \binom{3}{x}(\frac{1}{3})^x(\frac{2}{3})^{3-x}\binom{4}{4-x+0}(\frac{1}{2})^{x-0}(\frac{1}{2})^{4-x+0}\\=\binom{3}{0}(\frac{1}{3})^0(\frac{2}{3})^{3-0}\binom{4}{4-0+0}(\frac{1}{2})^{0-0}(\frac{1}{2})^{4-0+0}  \quad (x_1=x_2=0)\\+\binom{3}{1}(\frac{1}{3})^1(\frac{2}{3})^{3-1}\binom{4}{4-1+0}(\frac{1}{2})^{1-0}(\frac{1}{2})^{4-1+0}\quad(x_1=x_2=1)\\+\binom{3}{2}(\frac{1}{3})^2(\frac{2}{3})^{3-2}\binom{4}{4-2+0}(\frac{1}{2})^{2-0}(\frac{1}{2})^{4-2+0}\quad(x_1=x_2=2)\\+\binom{3}{3}(\frac{1}{3})^3(\frac{2}{3})^{3-3}\binom{4}{4-3+0}(\frac{1}{2})^{3-0}(\frac{1}{2})^{4-3+0}\quad(x_1=x_2=3)$
Ok, you can add up them to get the probability.
And remember for binomial distribution the random variable is total number of success. In your case $X_1$ and $X_2$ can be both 0, 1,2,3 for the total number of success of the two distribution.This answer in fact the same as the answer by Daniel and comment by ACE.
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