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 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 $1 \le X \le 4$.
Note:
$$\begin{align*} \Pr[X = 0] &\approx 0.107374 \\ \Pr[1 \le X \le 4] &\approx 0.859832 \\ \Pr[5 \le X \le 10] &\approx 0.0327935. \end{align*}$$
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.