In an examination, there are $5$ multiple select questions (one or more options can be correct) and each question has four options. Amanda decided to mark the answer at random. Find the probability that she gets exactly two questions correct (assume that the answer to each question is independent of the other questions).
For this applying the Binomial formula: $P(E) = {n \choose k} p^k (1-p)^{ n-k}$
$$\text{Correct answer is} = \frac{5!}{2!×3!}(\frac{1}{15})^2(\frac{14}{15})^3$$
How can I arrive at $p = \frac{1}{15}$? Please help me understand.
The reason why $p = 1/15$ is because you are told that "one or more options can be correct." If each question has four options, say A, B, C, D, then there are $2^4 - 1 = 15$ ways to select some non-empty subset of these four options. For example, you could select B and C as the answer. Or you can select all four as the answer. The only choice you cannot make is one where none of the options is selected, which is the reason for subtracting $1$. It follows that if Amanda marks an answer at random, and we assume that any of the possible answer choices are equally likely, the probability she answers the question correctly is $p = 1/15$.