Let there be a discrete random variable s.t. $$X \sim \text{Binom}(20,0.02)$$ and $X(\Omega) = \{0,1,2,\ldots,20\}$ Let there be also a constant $C$ s.t. $$\Pr(X\leq\frac{120}{C}) > 0.99$$ What is the maximum $C$ such that the above relation holds? And how should this problem be solved in R?

What has already been known

pbinom(4, size=20, prob=0.02)  

gives the probability that $X$ is less than 4.


Do you know the function qbinom? This gives the quantile of the binomial distribution.

We have that qbinom(0.99, size = 20, prob = 0.02) = 2, so $$ \mathbb{P}(X \leq 2) = 0.99. $$

So the maximum $C$ for which $$ \mathbb{P}\mathopen{}\left(X \leq \frac{120}{C}\right) = 0.99 $$ holds is $C=60$.


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