Mary can answer 20/25 problems correct… simple probability

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.

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

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

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

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

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How many questions are on the midterm? Just the 10? – Jed Mar 13 '14 at 23:01
Isn't the probability that Mary will know the first problem 20/25 (and not 10/20)? – Jed Mar 13 '14 at 23:05
I didn't really know which should be used regarding 20/25 or 10/20, wasn't sure it would make a difference... :-/ – CODe Mar 13 '14 at 23:28

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 $$P(X\ge8)$$

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I appreciate the effort, but we're not quite to hypergeometrics yet, could you perhaps explain what about my answer (using the way it was formatted) was incorrect? I'd like to learn from my mistake. – CODe Mar 13 '14 at 23:30
@CODe Ok, then. Why do you say that there is 10/20 to answer correct the first one? That is incorrect. If you do not use hypergeometric, then you need combinations. Have you done combinations? – Jimmy R. Mar 13 '14 at 23:34
I see, so instead of 10/20 it should have been 20/25, then 19/24, etc for each correctly answered problem? At the time this answer was jotted up, we hadn't done combinations, and I wanted to see what I had done wrong in my thought process, since that will hopefully correlate correctly when we start using combinations more. – CODe Mar 13 '14 at 23:40
@CODe Yes, that looks better. When you start to use combinations you will get the result faster. – Jimmy R. Mar 13 '14 at 23:50
Thanks for taking the time to answer my follow-ups, I know once you start using the shortcuts it's difficult to go back, I appreciate you answering in a way I could easily understand. :) – CODe Mar 13 '14 at 23:55