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Anne is running a rubber duck shooting stall at the local fate. She observes that her latest player, Mark, who is a keen shooter, has only 0.15 chance of missing the shot.

Anne wonders what the probability is that Mark will need to take at least 3 shots until he has a shot where he doesn't miss.

My working,

$\begin{align} Pr(X \geq 3) &= 1 - Pr(X \leq 2)\\ &=1 - [Pr(X = 1) + Pr(X = 2)]\hspace{1cm} \text{(since the number of shots can only be 1 or 2)} \\&= 1 - (0.15)^0(0.85)^1 -(0.15)^1(0.85)^1 \hspace{1cm} \text{(by geometeric distribution formula)}\\& = 0.0225 \end{align}$

Hence, the probability that Mark will need to take at least 3 shots until he has a one where he doesn't miss is $2.25\%$.

This answer feels correct, but I'm not sure about where I omitted $Pr(X = 0)$. Can I do this?

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Random variable $X$ is the number of shots needed to get the shot where Mark doesn't miss. The distribution if $X$ is geometric. You noted that

the number of shots can only be 1 or 2

So $Pr(X=0)=0$, as well as $Pr(X=-5)$ or $\Pr(X=-150)$. Do we need to take the last two probabilities into account?

The other way to find the probability that Mark will need to take at least 3 shots until he has a one where he doesn't miss is to note that he should have two shots where he misses. Really: if he misses in first two shots then he needs in third shot. And vice versa: if at least three shots are required, he miss in first two shots.

Therefore the required probability is $0.15^2=0.0225$.

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  • $\begingroup$ @Retty With pleasure! $\endgroup$ – NCh Apr 16 '17 at 10:40

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