Mary can answer 20/25 problems correct... simple probability

Question

Question: A teacher gave his class $25$ problems and told his students that he would select $10$ of them to put on their midterm. Mary can figure out how to answer $20$ of the problems, what is the probability that Mary will be able to answer at least $80\%$ of the questions on the exam?

Answer Explanation: My answer for this deals with assuming Mary can either get a $100 (10/10)$, a $90 (9/10)$, or an $80 (8/10)$, so I wrote a probability for each and added them together. Since she can correctly answer $20$, I know that the first problem she has a $10/20$ chance of getting it correct, then a $9/19$ since we remove one, etc. I also know she has a $5/25$ chance on the first problem of getting it wrong, and $4/24$ for the next one, etc.

Answer:

$P(100) = \text{no problems incorrect} = (10/20) * (9/19) * (8/18) * (7/17) * (6/16) * (5/15) * (4/14) * (3/13) * (2/12) * (1/11) $

$P(90) = \text{1st problem incorrect} = (5/25) * (10/20) * (9/19) * (8/18) * (7/17) * (6/16) * (5/15) * (4/14) * (3/13) * (2/12)$

$P(80) = \text{1st two problems incorrect} = (5/25) * (4/24) * (10/20) * (9/19) * (8/18) * (7/17) * (6/16) * (5/15) * (4/14) * (3/13)$

Thus... $P(>= 80\%) = P(100) + P(90) + P(80)$

Why is my answer incorrect?

Answer 1

Hint: The number $X$ of answers that Mary can answer correctly in the exam is a hypergeometric random variable with parameters $N=25$ (the population of the questions), $K=20$ (the questions that Mary can answer) and $n=10$ (the 10 questions that will be choosen in random out of the $25$ for the exam). In symbols $$X \sim \mathrm {Hypergeometric} (N=25,K=20,n=10)$$ As you have correctly thought, you want to calculate the probability $$P(X\ge8)$$

Answer 2

Here's a hint at a different way of doing it. The prof has 20 'good' questions to choose from (which Mary can do) and 5 'bad' questions which she can't. Let $A$ be the number of possible exams the prof could set, and $B_n$ be the number of possible exams containing exactly $n$ bad questions.

Then the required probability is $\frac{B_0+B_1+B_2}{A}$

$A$ will be equal to '25 choose 10' $B_1$ will be equal to '5 choose 1' x '20 choose 9', and you can find similar expressions for $B_0$ and $B_2$.

Quoting the general fact that $\text{(n choose k)}=\frac{n!}{k!(n-k)!}$, we can compute these quantities explicitly and hence the answer.