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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 $k$?

Thanks in advance!

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  • $\begingroup$ 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? $\endgroup$
    – JMoravitz
    Apr 3, 2017 at 21:36
  • $\begingroup$ 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. $\endgroup$
    – JMoravitz
    Apr 3, 2017 at 21:40
  • $\begingroup$ 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 $\sqrt{npq}$? $\endgroup$ Apr 3, 2017 at 21:53
  • $\begingroup$ Not a poisson distribution, a normal distribution. The normal distribution is the limit of a binomial distribution, not the poisson distribution. As for why the mean and standard deviation are what they are, these are well known properties of the binomial distribution. As to the question of whether or not it is sufficiently large... its "large enough" that it gives us at least a ballpark estimate, but small enough that we should check more closely if able. $\endgroup$
    – JMoravitz
    Apr 3, 2017 at 22:00
  • $\begingroup$ 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? $\endgroup$ Apr 3, 2017 at 22:11

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