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I'm confused on how you can interpret $n>2$. I want to use $\mathbb{P}[U>1] = 1 - (\mathbb{P}[U=1] + \mathbb{P}[U=0])$ but not sure how to implement $n>2$ into the formula.

Any ideas?

gt6989b
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Tingo Hugo
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  • It's not important, it's just there to make the question make sense, though even that only really needs $n>1$ at most. – Ian Dec 12 '22 at 00:51
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    If $n=2$ your answer would still be correct and would give $P(U>1)=1-((1-p)^2+2p(1-p))=p^2$ . If $n=1$ your answer would still be correct and would give $P(U>1)=1-((1-p)^1+p^1)=0$. If $n=0$ your answer would still be correct and would give $P(U>1)=1-((1-p)^0+0)=0$ though your answer for general $n$ and $p$ might have issues with the $+0$ there - you want to be clear that ${0 \choose 1}p^1(1-p)^{0-1}=0$ even when $p=1$. So strictly speaking the $n>2$ is unnecessary and does not affect the general answer, though $n>0$ might prevent pointless arguments. – Henry Dec 12 '22 at 01:05

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