Assume that i have an MCQ exam that consists of n questions where each question has k choices, what is the probability of passing this exam by picking answers randomly ?

I tried to solve this question with my very basic understanding of probabilities, so i want to know whether my solution is correct or not, i am also curious about knowing other solutions !

So my solution is the following:

the chance of picking the right answer is $\frac{1}k$ for a single question, so the chance of doing the same for j questions is $(\frac{1}k)^j$

Since in order to pass i have to correctly answer >= 50% of the exam

in this case it's possible that i correctly solve $\lceil\frac{n}2\rceil$ questions or $\lceil\frac{n}2\rceil + 1$ questions or .... or $n$ questions, as any of these scenarios will make me pass

So based on that, the probability of passing the exam should be equal fo this sum

$$ p = \sum_{i = \lceil\frac{n}2\rceil}^{n}(\frac{1}k)^i$$

So is this solution right ?

  • $\begingroup$ You did not take int account which questions you answered correctly or the fact that there are $k - 1$ ways to answer a question incorrectly. $\endgroup$ Commented Dec 16, 2021 at 12:26

1 Answer 1


For clarity, I will consider $n$ even and $n$ odd separately.

Let $p = (1/k), q = (1 - p).$

In general, the chances of exactly $r$ correct choices are

$$\binom{n}{r}p^r q^{n-r}.$$

$\underline{n ~\text{even} ~= 2s}$

Chance of passing equals

$$\sum_{r = s}^n \binom{n}{r} p^r q^{n-r}.$$

$\underline{n ~\text{odd} ~= 2s + 1}$

Chance of passing equals

$$\sum_{r = s + 1}^n \binom{n}{r} p^r q^{n-r}.$$


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