# What is the probability of getting at least 50 questions of 100 right?

If there are 100 MCQs with 4 options each. The probability that a person gets an question right is 0.25. What is the probability of getting at least 50 questions of 100 right?

• Pretty small; surely enough to convince the candidate that learning the subject would have been a good idea. For a numerical answer, I'd first try the normal approximation to the binomial distribution. – Lord Shark the Unknown Oct 20 '18 at 4:26

The number $$X$$ of questions correct if the student is answering strictly at random (with no knowledge of the subject matter) has $$X \sim \mathsf{Binom}(n = 100,\, p = 1/4).$$

So the probability of getting exactly fifty questions correct is $$P(X = 50) = {100 \choose 50}\left(\frac 14\right)^{50}\left(\frac34\right)^{50} = 4.5073 \times 10^{-8}.$$ Computation using R statistical software:

dbinom(50, 100, 1/4)
## 4.507311e-08


The probability of getting at least fifty correct is

$$P(X \ge 50) = \sum_{k=50}^{100} {100 \choose k}\left(\frac 14\right)^{k}\left(\frac34\right)^{100-k} = 6.6385 \times 10^{-8}.$$

1 - pbinom(49, 100, 1/4)
## 6.638502e-08


Both probabilities are very small because most of the probability in the distribution $$\mathsf{Binom}(n = 100,\, p = 1/4)$$ is centered near $$\mu = E(X) = np = 100(1/4) = 25.$$

Here is a figure that shows the distribution of $$\mathsf{Binom}(n = 100,\, p = 1/4)$$ along with the density function of $$\mathsf{Norm}(\mu = 25, \sigma = 4.33),$$ where $$\sigma = \sqrt{np(1-p)} =$$ $$\sqrt{75/4} = 4.3301.$$

Unless you are using software (or a statistical calculator) in your class, my guess is that you are supposed to use the normal approximation to the binomial distribution to approximate the very small value of $$P(X \ge 50).$$ (As @LordShark commented.) I will show the start of that procedure, and let you verify it and finish it for yourself:

$$P(X \ge 50) = P(X > 49.5) = P\left(\frac{X - \mu}{\sigma} > \frac{49.5 - 25}{4.3301} \right)\\ \approx P(Z > 5.658) = ?$$ where $$Z$$ is a standard normal random variable.

Hint: The number of ways to get $$n$$ questions right is $$\binom{100}{n}$$, and given any set of $$n$$ questions to get right the probability of doing so is

$$(0.25)^n(0.75)^{100-n}.$$

Can you solve it from there?

• Can the downvoter please explain what I should do to improve this answer? – Carl Schildkraut Oct 20 '18 at 5:50
• I haven't downvoted, but... This is the simplest way to figure out how to get exactly 50 questions right. Calculating the probability of at least 50 right with this method, though, is a real chore. – Arthur Oct 20 '18 at 6:49
• @Arthur I don't believe there's a better method to calculate it, is there? (Other than trying to get a numerical estimate, of course) – Carl Schildkraut Oct 20 '18 at 16:57
• I didn't downvote, but you found $P(X = 50)$ instead of $P(X \ge 50)$ and did not mention the assumption that the student is purely guessing. I'd guess the downvote focused on one of those issues. But this is not a bad start, so I'll cancel the downvote with a (+1). // There is no obligation to finish answering a hwk question, but if you don't, you should say it's only a start. – BruceET Oct 21 '18 at 21:30