# What is the probability of me passing an MCQ exam by picking answers randomly?

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 ?

• 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. Commented Dec 16, 2021 at 12:26

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

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