How to calculate the probability of one 1 customer in 100 making a pruchase, when the published statistic is 1 in 8 make a purchase? My college probability is more than a bit fuzzy. I've tried searching, but can't seem to find the right keywords to find what I need.  I also don't know if what i want to do is even possible.
Here's the scenario:


*

*I'm an affiliate marketer who has referred 100 new members to a website, and they sign up for a free membership. 

*The website claims that 1 in 8 free memberships convert to a premium membership.  I do understand that i won't necessarily have 12 conversions.

*I may not trust the website and would like to calculate the probability that out of 100 referrals I get 0 conversions, 1 conversion, 2, conversions, ...


Is this something that can be calculated given the information I have?  Any help or pointers would be greatly appreciated.
Thanks,
Paul
 A: By your point 2, we should assume that the probability that a given customer converts to a premium membership is $\frac{1}{8} = 12.5 \%$. Additionally, we can safely assume that all the decisions made by the customers are independent of each other.
Then, the answer to your question is given by a Bernoulli type formula. Here is the general formula: If you have $n$ customers, the probability that exactly k convert is 
$$ P(k \ out \ of \ n) = \frac{n!}{k!\cdot(n-k)!} \frac{1}{8}^k \cdot \frac{7}{8}^{n-k}. $$
For example, the probability of having zero purchases of 100 customers is then
$$ P(0 \ out \ of \ 100) = \frac{100!}{0!\cdot(100-0)!} \frac{1}{8}^0 \cdot \frac{7}{8}^{100} = \frac{7}{8}^100 \sim 0.00016 \% . $$
You cna easily calculate other cases with the general formula I gave.
A: We can use the binomial method,

It says,$$$$ let $p$ be the probability that an event will happen,$$$$then let $q $ be it not happening,$$$$if the recordings are done $ n$ times, $$P(\text{Said event will occur r times})=\binom{n}{r} (p)^r(q)^{n-r}$$

Try yourself or click below,

So probability of conversion ,$$p=\frac{1}{8}$$Probability of non-conversion,$$q=\frac{7}{8}$$For 100 people,$$P(\text{getting n conversions})=\binom{100}{n} (\frac{1}{8})^n(\frac{7}{8})^{100-n}$$$$P(\text{n conversions})=\frac{7^{100-n}\binom{100}{n}}{8^{100}}$$

