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The question seems simple but I just can't solve it.

I have twelve test scores and three are to be picked at random to determine the overall test grade. How can I calculate the probability of my final grade based on three randomly picked tests from the pool?

The test scores are as follows:

100=4
33=2
66=6

100 33 66 66 66 100 66 66 100 66 33 100

Would running a simulation be the simplest solution? Is there a good solution?

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  • $\begingroup$ Is this for a class assignment? $\endgroup$ – gung - Reinstate Monica Apr 14 '14 at 15:22
  • $\begingroup$ No, I happen to be the only one in the office that has a BS instead of a BA and have taken any math/statistics related course in the last 10 years. All quantitative issues are sourced to me. Actually this was a question posed by my co-worker who is worried about his grades in a Masters program. I am simply interested in how this can be solved. I ended up running a simple simulation in Excel 10,000 times to get the probabilities. I am wondering whether there is a better way to do this instead of a brute Excel method. $\endgroup$ – James Apr 15 '14 at 0:15
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If you don't need an exact figure, simulation is, as you've discovered, easy to do and a good approach to getting an answer.

But if you wanted something more exact, you could write yourself a short program to iterate through all $\frac{12!}{9!} = 1320$ ways three test scores could be chosen. Each triple is equally likely to represent your final grade as any other, so the probability of a final grade being $X$ is the portion of triples with average grade $X$.

Finally, if you don't feel like writing any code, you could power through a smaller number of cases and model the sampling as a multivariate hypergeometric distribution. With this distribution you can calculate the probability of your final grade consisting of $n_{100}$ 100s, $n_{33}$ 33s, and $n_{66}$ 66s as

$$\frac{\binom{4}{n_{100}}\binom{2}{n_{33}}\binom{6}{n_{66}}}{\binom{12}{3}}$$

and go from there.

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  • $\begingroup$ 2nd paragraph: Shouldn't you divide by 3! as well, yielding 220 combinations (not permutations)? $\endgroup$ – rolando2 Apr 19 '14 at 13:56
  • $\begingroup$ There are indeed 220 combinations, and you could iterate over them if you like, too. Whichever you find easier to write your program for, I suppose. $\endgroup$ – BeyondTheZero Apr 21 '14 at 15:32

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