# Probability problem - Binomial distribution?

I was trying to solve the following problem:

"100 students registered for an exam. The professor knows that each of them will actually take the exam, independently from all the others, with probability 60%. How many tests does he have to print out so that the probability of having enough tests is 98%?"

I was thinking that calculating the probability of printing out enough tests for the exam, let's say $k$, is equivalent to calculate the probability that $k$ students will actually take the exam. So I thought this process can be modelled with a binomial variable, where each $X_i$ represents a student.

Then $$98=\mathbb P(X> k)=1- \mathbb P(X\le k)$$ $$\Leftrightarrow 0.02=\mathbb P(X\le k)=\binom{100}{k}\left(\frac{2}{5}\right)^k\left(\frac{3}{5}\right)^{100-k}=\binom{100}{k}\left(\frac{2}{3}\right)^k\left(\frac{3}{5}\right)^{100}.$$

Is this "one" correct way to proceed? If so, How can I solve this equation with respect to $k$?

• That would be fine once you fix some things, but perhaps a more convenient to calculate approach would be using normal distributions. $n=100,p=0.6,q=0.4$, the average number of students arriving will be $np=60$ and the standard deviation will be $\sqrt{npq}=\sqrt{24}\approx 4.899$. How many standard deviations above the mean (what z-score) will result in $98\%$ of the data lying to the left? How many tests does this method suggest then that you need? Now that you have a good estimate, you can check against the actual distribution itself, does it work as you hoped? Apr 3, 2017 at 21:36
• Also, note: $\binom{100}{k}(\frac{2}{5})^k(\frac{3}{5})^{100-k}$ is the probability $Pr(X\color{red}{=}k)$, not $Pr(X\color{red}{\leq}k)$, you'll have to add up terms to account for that, making the method described in my first comment that much more convenient. Apr 3, 2017 at 21:40
• So, are you using a Poisson variable? If so, I read it is supposed to be the limit of a Binomial distribution whith parameter $\lambda=np$. But $n$ is to be sufficiently big. Is this the case in your opinion? Moreover why is the standard deviation equal to $\sqrt{npq}$? Apr 3, 2017 at 21:53
• Perfect, I didn't know that a binomial variable could be approximated by a normal one for $n$ sufficiently large. Now I get what you meant. But one last question. What do you mean when you say "check again the actual distribution"? Am I supposed to use the estimate of $k$ in the formula I wrote above? Apr 3, 2017 at 22:11