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Math Help - Exam Probability

  1. #1
    Junior Member shannu82's Avatar
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    Exclamation Exam Probability

    A student is preparing for an upcoming exam. The professor for the course has given the class 30 questions to study from and plans to select 10 of the questions for use on the actual exam. Suppose that the student knows how to solve 25 of the 30 questions.

    a) What is the probability that the student will get perfect on the test?
    b) What is the probability that the student will get at least 8 questions correct on the test?

    Any help would be appreciated.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Find the probability that he knows every question:

    a) P(gets perfect) = 25/30 x 24/29 x ... x 16/21

    b) P(8 correct) = 25/30 x 24/29 x ... x 18/23 x 5/22 x 4/21
    Last edited by Unknown008; October 31st 2010 at 11:24 AM. Reason: Typo about the sign... sorry :o
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  3. #3
    Guy
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    Quote Originally Posted by Unknown008 View Post
    Find the probability that he knows every question:

    a) P(gets perfect) = 25/30 + 24/29 + ... + 16/21

    b) P(8 correct) = 25/30 + 24/29 + ... + 18/23 + 5/22 + 4/21
    This can't be right. The first two terms in P(gets perfect) sum up to more than 1. I think you meant write it as a product. I think (a) is correct if you replace the sum with a product, whereas (b) is not correct.

    The easy way to do this problem is with the hypergeometric distribution. See the sidebar on the right for the pmf.
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Oops, yes, I made a very bad typo. Thanks for pointing it out!
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  5. #5
    Guy
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    I think part (b) is still wrong. You forgot to count the number of ways he can get 8 right. What you have listed is the probability of getting the first 8 right and the last 2 wrong.
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  6. #6
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    For the a) part it is \dfrac{\binom{25}{10}\binom{5}{0}}{\binom{30}{10}}.

    For b) do that for 8, 9, & 10. Then add.
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  7. #7
    Junior Member shannu82's Avatar
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    Thank You!
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  8. #8
    MHF Contributor Unknown008's Avatar
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    Darn, I must have been tired last night...

    Yes, the shorter way is to use combinations.

    Of 30 questions, the professor had to choose any 10 of the 25 questions that the student knew, that is 25C10.

    The total possible combinations are 30C10.

    So, probability of getting all correct becomes:

    P(all\ correct) = \dfrac{^{25}C_{10}}{^{30}C_{10}}

    For the second one, the same thing applies, but I didn't see 'at least' yesterday night...

    P(at\ least\ 8) = \dfrac{^{25}C_{8} . ^5C_2}{^{30}C_{10}} +  \dfrac{^{25}C_{9} . ^5C_1}{^{30}C_{10}} +  \dfrac{^{25}C_{10} . ^5C_0}{^{30}C_{10}}

    Sorry again
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