Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

After completely guessing on quite a few tests (long story here), I've always wanted an accurate estimate of my scores, but I could never apply the probability theorems I learned in high school. I've never taken a formal course pre or post calculus statistics (nor have I taken calculus), but I find great curiosity in problems like these. It's totally fine if you use ridiculously hard math to explain, because I believe that someday I will come back and look at this problem again.

Here's the basic problem, and following are a bunch of variations that could be added to the problem.

Let's say I have 60 multiple choice questions each with 4 choices and that the total score is 100(%). Now, 20 of those questions are 3 points each, and 40 are 1 point each. Let's say that answer distribution is equal among the four choices (A,B,C,D). How would I find the probability of getting a score n, given that I randomly selected choices?

Variations: How would I calculate the probability of n if answer distribution was not equal, but normally distributed? If so, how would I calculate the probability of n if I chose equal answers, or in general, answers with restrictions?

This might not be an effective problem to solve with known probability theorems; I don't know. This might be something common for people that are super whizzes at statistics; I don't know either. Please tell me what you know about it though. Thanks!

share|cite|improve this question
What is a normal distribution on 4 answers? – Dilip Sarwate Apr 3 '12 at 13:17
up vote 4 down vote accepted

Actually, this is a fairly standard way to solve this using basic methods: Let $X_i$ denote the random variable that says how many points you got on the $i$-th question. Assume the first 20 questions are 3 points each, so $X_1,\ldots,X_{20}$ are 3 with probability $1/4$, and 0 with probability $3/4$. Of course, $X_{21}, \ldots, X_{60}$ are 1 with probability $1/4$ and $0$ with probability $3/4$.

To find out whether $X := \sum_{i=1}^{60} X_i = n$, it is easiest to note that $Y_1 := \frac{1}{3} \sum_{i=1}^{20} X_i$ is binominally distributed and so is $Y_2 := \sum_{i=21}^{60} X_i$ (with appropriate parameters). Thus, $P(X = n) = \sum_{i=0}^n P(3Y_1 = i \wedge Y_2 = n-i) = \sum_{i=0}^n P(3Y_1 = i) \cdot P(Y_2 = n-i)$, which should be easily solvable using the binominal distribution. Note that the last transformation uses the fact that $Y_1$ and $Y_2$ are independent.

Actually, it might be more interesting to work out the expected value (the average number of points) here, which is much easier by using linearity of expectation: $\mathbb E \sum_{i=1}^{60} X_i = \sum_{i=1}^{60} \mathbb E X_i = 20 \cdot \mathbb E X_1 + 40 \cdot \mathbb E X_{21} = 20 \cdot 3 \cdot 1/4 + 40 \cdot 1 \cdot 1/4 = 25$.

share|cite|improve this answer
Ok, I'm pretty sure that will be helpful to me in the future. What branch of math would this lie in? – inkyvoyd Apr 3 '12 at 10:05
Probability distribution under probability theory. – VelvetThunder Apr 3 '12 at 10:13
To be honest, I'm somewhat surprised that you did not learn the relevant stuff in school - these results are fairly basic, and I'm pretty sure almost all of this was covered when I was back in school (might be national differences at work here). – Johannes Kloos Apr 3 '12 at 10:43
I wish I had learned these things in high school -- it would have made my studies easier. Unfortunately, in Polish schools, the concept of a random variable is not introduced. Perhaps this is also the case where the OP studies? – user23211 Apr 3 '12 at 13:22
I actually don't understand half of these symbols. I am in freshman year of high school, and although my math is "advanced", I have not learned much about probability. – inkyvoyd Apr 3 '12 at 18:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.