# Binomial Random Variable Intuition

I have the following question from a stats course about binomial distribution:

A multiple choice test has 10 questions, each with 5 possible answers, only one of which is correct. A student who did not study is absolutely clueless, and therefore uses an independent random guess to answer each of the 10 questions.

Here, $$X$$ is the binomial random variable equal to the number of right answers. $$n = 10$$ and $$p = 0.2$$. I am struggling to work my intuition about a specific case. We have,

• Probability that at least one answer is correct is:

$$P(X\ge1) = 1 - P(X=0)$$ $$= 1 - 0.8^{10}$$ $$= 0.892$$

• Probability that the student will get at most 4 questions right: $$P(X\le4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$$ $$= 0.967$$

What I don't quite understand is, why does the first case have a lower probability that the second case? Isn't the first case relatively more inclusive (larger cardinality of events) and thus have a higher chance of occurring than the second case?

• The first excludes $P(X \le 0)$ while the second excludes $P(X \ge 5)$. Knowing $E[X]=2$, it is not particularly surprising that $P(X \le 0) > P(X \ge 5)$ – Henry Apr 23 at 16:14

The two events $$X \ge 1$$ and $$X \le 4$$ are not nested. That is to say, the first event excludes $$X = 0$$ and includes $$X > 4$$, whereas the second includes $$X = 0$$ and excludes $$X > 4$$. The set of outcomes these events have in common is $$1 \le X \le 4$$.
\begin{align*} \Pr[X = 0] &\approx 0.107374 \\ \Pr[1 \le X \le 4] &\approx 0.859832 \\ \Pr[5 \le X \le 10] &\approx 0.0327935. \end{align*}
Thus, the reason why the first probability is smaller than the second is because the event $$X = 0$$ is more likely than $$5 \le X \le 10$$; and this makes sense because with only a $$1/5$$ probability of answering a given question correctly, it is highly improbable to answer many of them correctly by random chance.
It can help to imagine what happens in limit case. If number of possible answers is very large, it's almost impossible to get even a single one right - so $$P(X \geqslant 1)$$ will be approximately $$0$$. And simultaneously it will be almost certain that student will not get more then $$4$$ answers right, so $$P(X \leqslant 4)$$ will be approximately $$1$$.
Sure at the first case you add more probabilities. But you have to regard the sknewness of the distribution as well. Almost the whole mass of the distribution is conentrated on $$P(X\leq 4)$$. Especially $$P(X=0)$$ is larger than $$P(X\geq 5)$$