# 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

## 1 Answer

Sometimes it helps to really write out what the scenario actually means. In other words, if the player does not miss until his 10th shot then we know that he must make 9 shots in a row and then miss on the 10th. If we know that each shot has an 80% chance of going in and a 20% chance of missing then:

$$P(\text{makes 9 in a row}) = P(\text{makes 1st}) \times P(\text{makes 2nd}) \times \dots \times P(\text{makes 9th}) = (0.8)^9$$

and then the probability he misses the 10th shot is 20%, i.e. 0.2. So all together we have

$$P(\text{doesn't miss until 10th}) = P(\text{makes 9 then misses}) = (0.8)^9 \times (0.2).$$

Hope that helps!

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