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


1 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.

  • $\begingroup$ Hi, thank you for your input. Could you please elaborate on how did you make the transition from the last summation? $\endgroup$
    – Orokusaki
    Feb 3, 2021 at 21:19
  • $\begingroup$ 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. $\endgroup$ Feb 3, 2021 at 21:36
  • $\begingroup$ 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 $\endgroup$
    – Orokusaki
    Feb 3, 2021 at 22:08
  • 1
    $\begingroup$ This is just the binomial formula. $\endgroup$ Feb 3, 2021 at 22:12

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