I am writing a paper for math class and have run into an issue. I am trying to determine the probability of getting $9$ (or more) kills out of $21$ hits in a volleyball game in a certain spot by using a binomial distribution function.
For $p$ I am using a probability determined from data I collected ($\frac{45}{66}$ is the probability I found of getting a kill in zone $1$).
There are 6 zones and I have observed the probability of getting a kill in each of the 6 zones. For x trials, I am using 21 and n successes (kill). When I plug in the numbers I am getting crazy answers, for example, getting at least 9 kills out of 21 attempts in zone one using a probability of a kill at $\frac{45}{66}$, that probably is coming up as $99$%... while some of the other zones where the probability is $.158$ probability of getting $9$ or more kills out of $21$ attempts come up as $2$%?
This can’t be right, can it? My thinking is that a binomial distribution function would work as there are only two options (kill or receive) and I have the observed probabilities for $p$. Maybe I need to rethink my strategy? Help would be appreciated.
If $p$ is the probability of a kill and $q = (1-p)$ and if each bernoulli trial [kill or no kill] is independent of the others, then the probability of getting exactly $k$ kills in $n$ trials is
$$\binom{n}{k}p^k q^{(n-k)}.$$
This means that if you want 9 or more kills out of 21, you want
$$\sum_{k=9}^{21} \binom{21}{k} p^k q^{(n-k)} ~~\text{where}~~ p = \frac{45}{66}.$$