# Simple geometric distribution (solution verification)

The question is:

In a hockey competition, a player scores $80\%$ of his shots. What is the probability that the player will not miss until his $10^{th}$ try?

So I did the following

\begin{align} &P(X) = (1-p)^{k-1}p \\ &P(.8) = (1-.8)^{9}(0.8)\\ &P(.8) = \frac{4}{9765625} \end{align}

Is this correct? Because my teacher has done: $(0.8)^{9}(0.2)$ which doesn't make sense since the formula states $(1-p)$.

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Hmmm... Please give which events have a probability .2 and .8 ? –  Mr.ØØ7 May 9 '13 at 11:34
In your case, $p$ is the probability of missing the shot, which is $(1-0.8)$ if $0.8$ is the probability of getting the shot. –  Michael Greinecker May 9 '13 at 15:27

$$P(\text{makes 9 in a row}) = P(\text{makes 1st}) \times P(\text{makes 2nd}) \times \dots \times P(\text{makes 9th}) = (0.8)^9$$
$$P(\text{doesn't miss until 10th}) = P(\text{makes 9 then misses}) = (0.8)^9 \times (0.2).$$