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!