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#### eatinbyzombies3

##### New member

- Jun 5, 2013

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thank you and I promise this is the last question today.

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- Thread starter
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- Jun 5, 2013

- 3

thank you and I promise this is the last question today.

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- Jan 26, 2012

- 4,052

Hopefully someone else will confirm this approach, but it seems like you can interpret this to be a binomial random variable. If not then my apologies in advance. Let's say that $P[\text{gold}]=0.65$.

The general formula for a binomial random variable is:

\(\displaystyle P[X=k]=\binom{n}{k}p^{k}(1-p)^{n-k}\)

where $p$ is the success probability, $n$ is the number of trials and $k$ is the number of successes. Can you fill in the information?