Here is a light question:

Say that my professor says she will choose 12 definitions from a list of 40, and I will have to answer 7 of those 12 on a test. Say that I know 24 of the 40 possible definitions she will choose from. What is the probability that at least 7 of the definitions on the test will be from the 24 I know?

This is really the case for my History Final Today. It did spark my curiosity as to how the probability would be calculated. I know binomial coefficients are involved, but any insights towards a complete solution?


For $0 \le k \le 12$, the probability that she chooses exactly $k$ of those $24$ that you know is $$ {24 \choose k} {16 \choose 12-k}\left/ {40 \choose 12} \right.$$ You must add these for $k$ from $7$ to $12$.

But I think it would be more productive to learn some of the other $16$ definitions.

  • $\begingroup$ Cool thanks! And yeah, obviously It would be ideal to learn all 40, but I only had access to 24. $\endgroup$ – Chezky Steiner May 23 '17 at 18:13

There are ${40}\choose{12}$ possible exams.

We need $n = 7..12$ correct solutions, and can make $12-n$ mistakes.

$\sum_{n=7}^{12} {{24}\choose{n}} {{16}\choose{12-n}}$

Divide for the ratio, take normal distribution approximation for practical use.


I'm not sure how to read the question. In one interpretation the $40$ seems to be irrelevant. You know she will choose her $12$ from $24$. There are $W = \binom{24}{12}$ ways she can do that.

Counting the ways you can guess $7$ or more you get $$ C = \binom{12}{7} + \binom{12}{8} + \binom{12}{9} + \binom{12}{10} + \binom{12}{11} + \binom{12}{12} . $$ The probability is $C/W = ...$

The other interpretation is that she will definitely use at least $1$ from the $24$ you know about. The probability then is ...

Rereading, there's a third interpretation that's probably what you mean. You know the answer for $24$ of the questions. Then @RobertIsrael 's calculation tells you the probability.

If it's not too late, take his advice and study them all.

  • $\begingroup$ yes, the third interpretation is what I meant. $\endgroup$ – Chezky Steiner May 23 '17 at 18:12

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