Probability: Finding n and p.

I am preparing for my finals and came across this question:

$\displaystyle \text{Let p be the probability that the vaccine is a success in any patient.}$

$\displaystyle \text{Find the sample size n and p required to be 95 percent assured that when the}$ $\displaystyle \text{vaccine is given to n infected patients, at least 1000 successful vaccinations will result}$

I am quite confused on this one since we have two unknowns. I tried using binomials distribution with $\displaystyle P(X \geq 1000)\;=\;1-P(X=1000)\;=\;0.95$ with n being the number of patients to be vacinated in order to get at least 1000 successes, but i cannot calculate both $\displaystyle n$ and $\displaystyle p$ from the same equation.

is there any other approach to solve this problem? thank you.

Re: Probability: Finding n and p.

Quote:

Originally Posted by

**chutiya** I am preparing for my finals and came across this question:

$\displaystyle \text{Let p be the probability that the vaccine is a success in any patient.}$

$\displaystyle \text{Find the sample size n and p required to be 95 percent assured that when the}$ $\displaystyle \text{vaccine is given to n infected patients, at least 1000 successful vaccinations will result}$

I am quite confused on this one since we have two unknowns. I tried using binomials distribution with $\displaystyle P(X \geq 1000)\;=\;1-P(X=1000)\;=\;0.95$ with n being the number of patients to be vacinated in order to get at least 1000 successes, but i cannot calculate both $\displaystyle n$ and $\displaystyle p$ from the same equation.

is there any other approach to solve this problem? thank you.

X ~ Binomial(n, p)

Pr(X < 1000) = 0.05

You will now need to use the normal approximation to make further progress.

Re: Probability: Finding n and p.

Thank You. Thats what I was thinking about.

Apparently, the value of p had been missing in the question, which was p=90%(which was the major source of my trouble) and I was supposed to find "n" only. That makes life easy. I just used normal approximation $\displaystyle np-1.64\sqrt{np(1-p)}=1000$ to find n.