Probability in multiple choice exams

I had this question once pop up in my math exam and I was never given the answer. Suppose a person is doing a multiple choice exam where he knows none of the answers. He has 5 answers to choose from so the probability of him choosing the right one is 20%. Given that he has to answer 40 questions randomly what is the probability of him passing the exam if the passing mark is a 40%. (note each question holds 2.5%). Please let me know the method that you used and steps, I used binomial distribution but i got it wrong for some reason.

P(X ≥ 16) = .002936 or approx. .3%

Keep in mind that we need P(16 *or more* correct), not just P(exactly 16 correct). This could possibly have been your error, since it is a very common error.

Yes, the exact distribution to use is the binomial, but to calculate this probability, you would need to use the normal approximation to the binomial because the numerous binomial distribution calculations would be extremely laborious.

Let X = number of correct answers

Note that X has mean mu = np = 40(0.2) = 8, and standard deviation
sigma = sqrt(40*0.2*0.8) = sqrt(6.4).

To obtain a somewhat more accurate normal approximation, you can use a continuity correction by associating each number n of correct answers with the interval [n - 0.5, n + 0.5]. The reason for this correction is that the binomial is discrete while the normal is continuous.

Since Z = (X - mu)/sigma, the normal approximation for P(16 or more right) is

P(X >= 15.5) = P(Z >= (15.5 - 8)/sqrt(6.4))
= P(Z >= 2.96)
= 1 - P(Z < 2.96)
= 1 - 0.9985, from the normal table
= 0.0015 .

Not surprisingly, the student is very unlikely to pass the exam by just guessing.

Lord bless you today!

First, you have to know how many questions the person has to get correct to pass

40 /2.5 = 16 questions

P(passing) = P(guessing 16 correct) + P(guessing 17 correct)+.. +P(guessing 40 correct)