I have the following question in one of my tutorial.
Background: A manufacturing company developed 40000 new drugs and they need to be tested. Question The QA checks on the previous batches of drugs found that — it is four times more likely that a drug is able to produce a better result than not. If we take a sample of ten drugs, we need to find the theoretical probability that at most 3 drugs are not able to do a satisfactory job.
I think we need to use the cumulative probability distribution $F(3) = P(X\lt 3).$ However not sure how to calculate it. Any guidance is helpful.
If you let $G$ be the event that a drug produces better results and you let $B$ be the event that a drug produces not as good of results, then the first step in your question is to solve $4P(B)=P(G)$ with the condition that $P(B) + P(G)=1$. From this, we get $P(B)=\frac{1}{5}$ and $P(G)=\frac{4}{5}$.
Accordingly, if you let $X$ be the event that out of 10 trials, you pick $x$ amount of drugs that do not produce as good of results as the other drug, then $X$ is a random variable with a Binomial Distribution, where $n=10$ and $p=1/5$.
So, $P(X=x)={10 \choose x}\frac{1}{5}^{x}\frac{4}{5}^{10-x}$, for $0 \leq x \leq 10$.
All that is left is to compute $P(X \leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)=\sum_{i=0}^{3}{10 \choose i}\frac{1}{5}^{i}\frac{4}{5}^{10-i}$