Sum of independent Binomial and Geometric distributions

Question

This is a question from a mock test in my Intro to Probability coure:

Let $X$~$Bin(26,\frac{1}{23})$ and $Y$~$Geo(\frac{1}{2})$ be two independent random variables. Let $Z=X+Y$. Calculate $\Bbb{P}[Z=46]$.

My initial solution:

$\Bbb{P}[Z=46] = \Bbb{P}[X+Y=46] = \sum_{k=0}^{26}{\Bbb{P}[X+Y=46, X=k]} = \sum_{k=0}^{26}{\Bbb{P}[Y=46-k]\cdot\Bbb{P}[X=k]} = \sum_{k=0}^{26}{(\frac{1}{2})^{46-k}\binom{26}{k}(\frac{1}{23})^{k}(\frac{22}{23})^{26-k}} $

I got stuck calculating this sum. If anyone could help me, it would be much appreciated.

Thank you

Answer 1

You are very close to get the desired answer. Note the following: $$\sum_{0 \leq k \leq 26} (\frac{1}{2})^{46 - k} \binom{26}{k} (\frac{1}{23})^k (\frac{22}{23})^{26 - k} = (\frac{1}{2})^{20} \sum_{0 \leq k \leq 26} (\frac{1}{2})^{26 - k} \binom{26}{k} (\frac{1}{23})^k (\frac{22}{23})^{26 - k} = \frac{1}{2^{20}} \sum_{0 \leq k \leq 26} \binom{26}{k} (\frac{1}{23})^k (\frac{11}{23})^{26 - k} = \frac{1}{2^{20}} (\frac{1}{23} + \frac{11}{23})^{26} = \frac{1}{2^{20}} \frac{12^{26}}{23^{26}},$$ which can be further simplified to give $\dfrac{2^{32} 3^{26}}{23^{26}}.$

Digressing a little from your very particular example, the question of determining the distribution of the sum of two continuous random variables is a very interesting one and it is one of the places where the convolution of two functions proves itself once again to be quite an useful tool. For references, you might want to start by reading this wikipedia page.