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Let $X,Y,Z$ be independent discrete random variables with $X\sim\text{Bin}(6,7/8),$ $Y\sim\text{Bin}(5,7/8)$ and $Z\sim\text{Geo(1/4)}$. Futhermore let $M:=\max(X+Y,Z)$.

Compute $\Pr[M>10]$.

My approach so far based on the fact that $X+Y\sim\text{Bin}(11,7/8)$:

$$\Pr[M>10] = 1-\Pr[M\leq 10]=1-\Pr[X+Y\leq 10]\cdot\Pr[Z\leq 10]\\ \Pr[X+Y\leq 10] = 1-\Pr[X+Y=11]=1-(7/8)^{11}\approx 0.7698\\ \Pr[Z\leq 10]=1-(1-1/4)^{10}\approx 0.9437\\ \Longrightarrow \Pr[M>10]\approx 1-0.7698\cdot 0.9437=0.2735$$

However I think this cannot be true because I haven't considered $\max$ yet. Can anyone explain me what did I miss?

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+1 For showing all your work. I wish more homework questions were like this. –  Sasha Oct 3 '12 at 15:02
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up vote 3 down vote accepted

You already implicitly used the $\max$ in the last equality on the first line.

$1-\text{Pr}[M\leq 10]=1-\text{Pr}[\max(X+Y, Z) \leq 10]=1-\text{Pr}[X+Y\leq 10]\text{Pr}[Z\leq 10]$

The maximum of $X+Y$ and $Z$ is less or equal to $10$ is equivalent to each of them being less than or equal to $10$ since they are independent.

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