# Sum of independent Binomial and Geometric distributions

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

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 $$\dfrac{2^{32} 3^{26}}{23^{26}}.$$
• If I understand correctly, you are interested in the transition from $\binom{26}{k} (\frac{1}{2})^{26 - k} (\frac{1}{23})^k (\frac{22}{23})^{26-k}$ to $\binom{26}{k} (\frac{1}{23})^k (\frac{11}{23})^{26-k}.$ Here I did nothing more than multiplying $(\frac{1}{2})^{26 - k}$ with $(\frac{22}{23})^{26-k}.$ I hope I understood correctly, but if not do let me know. Feb 3, 2021 at 21:36
• Hi, I meant going from $\frac{1}{2^{20}} \sum_{0 \leq k \leq 26} \binom{26}{k} (\frac{1}{23})^k (\frac{11}{23})^{26 - k}$ to $\frac{1}{2^{20}} (\frac{1}{23} + \frac{11}{23})^{26}$. Thank you Feb 3, 2021 at 22:08