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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!

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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: $\mathbb E \sum_{i=1}^{60} X_i = \sum_{i=1}^{60} \mathbb E X_i = 20 \cdot \mathbb E X_1 + 40 \cdot \mathbb E X_{21} = 20 \cdot 3 \cdot 1/4 + 40 \cdot 1 \cdot 1/4 = 25$.

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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. – Quixotic 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

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