Given the percentage, what's the probability it will happen exactly? If a drug is effective $75\%$ of the time, what's the probability that it will be effective on EXACTLY $15$ out of $20$ people. 
Is there a formula or list of steps for this type of question? 
 A: Yes. 
The probability that it works on a specific set of 15 people, and also doesn't work on a specific set of $5$ people, is:
$$P_{\rm{specific\ 15\ people}} = 0.75^{15}\cdot 0.25^5$$
But of course, we need to update the number based on how many ways we can divide a group into sets of $5%$ and $15$. This is given by the binomial coefficient:
$${{20}\choose {15}}=\frac{20!}{15!\cdot 5!}$$
So all together we have that the final probability is the product of these two numbers:
$$P={{20}\choose {15}}\cdot 0.75^{15}\cdot 0.25^5\approx 0.2=20\%$$
A: This is given by the binomial distribution:
$P(X=15)={20 \choose 15}0.75^{15}(1-0.75)^{20-15}$
A: Ok, so the answer has been posted but not in a non-random variable way, here goes.
The first issue is to convert percentages into probabilities (decimals). They're easier to work with. If a drug works 75% of the time then if you give me a person, the chance it'll work on her is $0.75$. The chance it won't work (it has to do one or the other) is $0.25$.
The chance then that it will work on a specific 15 people - say the tallest 15 - will be $$0.75^{15}(1-0.75)^{5}$$
Then we just need to decide how many different ways there are of choosing those 15 people. This is given by ${20 \choose 15}$ '20 choose 15'. If you want an actual number then: ${{20}\choose {15}}=\frac{20!}{15!\cdot 5!}$
Hence, the chance it will work on exactly 15 is
$$P(X=15)={20 \choose 15}0.75^{15}(1-0.75)^{5}$$
