0
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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