I understand that there are only four more questions that he needs to answer correctly for him to past the test.I know the basic equation but I'm not sure what to do from there.

A student is writing a multiple choice test consisting of 40 questinons, each of witch provides 4 possible choices. He is certain that he has 16 questions correct. If he guesses for all of the remaining 24 questions, what is the probability that he will pass the test?

Hint: There are only 24 questions left to choose from.

  • 1
    $\begingroup$ what is the passing grade? $\endgroup$ – gt6989b May 10 '17 at 21:24
  • $\begingroup$ 50%, so if he just wants to pass then he only needs to answer 4 more questions correctly $\endgroup$ – Julia Fraser May 10 '17 at 21:27


Notice that the guess on each question (assuming no answers could be eliminated) gives him a 1/4 probability of being right. So, let $X$ be the number of questions he guesses correctly. Then, $X$ is Binomial with parameters $n = 40-16=24$ and $p = 1/4$.

Let $P$ denote the number of problems he needs to get correct on this test to pass. Then, $$ \mathbb{P}[\mathrm{passing\ grade}] = \mathbb{P}[X \ge P-16], \quad \mathrm{where} \quad X \sim \mathcal{B}(24, 1/4). $$ Can you finish this?

UPDATE Now use $P=20$ $$ \begin{split} \mathbb{P}[\mathrm{passing\ grade}] &= \mathbb{P}[X \ge P-16] = \mathbb{P}[X \ge 4]\\ &= 1- \mathbb{P}[X = 0] - \mathbb{P}[X = 1] - \mathbb{P}[X = 2] - \mathbb{P}[X = 3] \end{split} $$ Since $X \sim \mathcal{B}(24, 1/4)$, we have $$ \begin{split} \mathbb{P}[X = 0] &= \binom{24}{0} 0.75^{24}\\ \mathbb{P}[X = 1] &= \binom{24}{1} 0.25 \cdot 0.75^{23}\\ \end{split} $$ Can you finish this now?

  • $\begingroup$ i got 1.28^32... doesn't seem right... $\endgroup$ – Julia Fraser May 10 '17 at 21:43
  • $\begingroup$ @JuliaFraser please see the update $\endgroup$ – gt6989b May 10 '17 at 21:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.