Most Probable Average Marks

Question

There is a class of 40 students. A professor conducted a quiz on which consisted of 10 Multiple Choice Questions. Each question had 4 options, out of which only one is correct. Each question carried 1 mark and there was no negative marking. What is the most probable average marks of this class? (Considering that students have no clue about the questions and they do not copy from each other).

Answer 1

Start with $n$ i.i.d random variables $x_i \in \{0,1\}$. Let $p$ be the probability for $x_i$ to take value 1. The probability for $\sum_{i=1}^n x_i = k$ will be given by the formula $\binom{n}{k} p^k (1-p)^{n-k}$ .

Let $K$ be the most probable value of $k$.

By comparing the probability for the $K$ and $K+1$ term, we have:

$$\begin{align} & \binom{n}{K} p^{K} (1-p)^{n-K} \ge \binom{n}{K+1} p^{K+1} (1-p)^{n-K-1}\\ \implies & 1 \ge \frac{n-K}{K+1} \frac{p}{1-p}\\ \implies & K \ge (n+1)p - 1 \end{align}$$

By compariing the probability for the $K$ and $K-1$ term, we have:

$$\begin{align} & \binom{n}{K} p^{K} (1-p)^{n-K} \ge \binom{n}{K-1} p^{K-1} (1-p)^{n-K+1}\\ \implies & 1 \ge \frac{K}{n-K+1} \frac{1-p}{p}\\ \implies & (n+1)p \ge K \end{align}$$

This means in general, the most probable value of $k$ is sandwiched between $(n+1)p - 1$ and $(n+1)p$.

Back to our question. Since we have $40$ students, each is going to take $10$ questions, we have $n = 400$ and $p =\frac{1}{4}$. Furthermore, if $X$ is the average mark for the whole class, $k = 40 X$. This means the probability distribution for $X$ is given by: $$\binom{400}{40X} \left(\frac14\right)^{40X} \left(\frac34\right)^{400 - 40X}$$

Using above bound, we have:

$$99.25 = (n+1)p - 1\le K \le (n+1)p = 100.25 \implies K = 100 \implies X = 2.5$$

By brute force computation, the most probable average mark is indeed 2.500.

$$\begin{array}{c|c} \text{avg. mark} & \text{probablity}\\ \hline\\ 2.400 & 0.041900165340674\\ 2.425 & 0.043771994032869\\ 2.475 & 0.045871413522661\\ \color{red}{2.500} & \color{red}{0.046024318234403}\\ 2.525 & 0.045568631915251\\ 2.550 & 0.044526212230915\\ 2.575 & 0.042941136714604\\ 2.600 & 0.040876658987940\\ \end{array}$$