<s>[INST] Probability problem with binomial/multinomial distribution Mary knows the answers to $20$ of the $25$ multiple choice questions on the Psychology $101$ exam, but she has skipped several of the lectures, she must take random guesses for the other five. Assuming each question has four answers, what is the probability she will get exactly $3$ of the last $5$ questions right?
 [/INST] We need to make some assumptions. We will assume the $5$ questions she does not know the answers to are equally likely to be any $5$ of the $25$ questions. 
We want the probability she gets exactly $2$ of the last $5$ questions wrong. This can happen in various ways. The last $5$ questions may contain $2$ questions she has to guess on, and she gets both  wrong. Or the last $5$ questions may contain $3$ questions she has to guess on, and she gets $1$ right. Or they may contain $4$ questions she has to guess on, and she gets $2$ right. Or they may contain $5$ questions she has to guess on, and she gets $3$ right.
$2$ questions: The probability there are $2$ questions she does not know the answer to is $\frac{\binom{5}{2}\binom{20}{3}}{\binom{25}{5}}$. Given this, the probability she gets them both wrong is $\left(\frac{3}{4}\right)^2$. This gives probability $\frac{\binom{5}{2}\binom{20}{3}}{\binom{25}{5}}\left(\frac{3}{4}\right)^2$. 
$3$ questions: The probability there are $3$ questions she does not know the answer to is $\frac{\binom{5}{3}\binom{20}{2}}{\binom{25}{5}}$. Given this, the probability she gets $2$ of them wrong (so $1$ right) is $\binom{3}{1}\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^2$. Now we can write down the associated probability by multiplying.
$4$ questions, $5$ questions: It's your turn.
Now add up.  
</s>