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Math Help - Odds of passing a test.

  1. #1
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    Odds of passing a test.

    An instructor gives her class a set of 10 problems with the information that the final exam will consist of a random selection of 5 of them. If a student has figured out how to do 7 of the problems, what is the probability that he or she will answer correctly

    (a) all 5 problems?

    So the sample space, S is \binom{10}{5}=252

    S_5 will be the event that the student doesn't know all 5 problems. S_4 that he doesn't know only 4 questions, and so on.

    S_0=S-(S_1+S_2+S_3)

    Since the student knows 7 problems the event that the test contains more then 3 unknown problems has to be a null set as there are at most 3 unknown problems.

    OK so there are two pools of problems, the seven that he knows and the 3 that he doesn't. So \binom{7}{n_1} and \binom{3}{n_2} where n_1+n_2=5 and only four valid combinations that sum to five, (n_1,n_2)=\{(2,3)(3,2)(4,1)(5,0)\}. So,

    \binom{7}{2}\binom{3}{3}+\binom{7}{3}\binom{3}{2}+  \binom{7}{4}\binom{3}{1}-231

    So S_0=252-231 and \frac{21}{252}=0.0833\overline{3}

    This is the listed answer but does this process look good/correct?

    Also \frac{\binom{7}{2}}{\binom{10}{5}}=0.0833\overline  {3}

    (b) at least 4 of the problems?

    So this should just be
    S_0+S_1=S-(S_2+S_3)

    252-\binom{7}{2}\binom{3}{3}+\binom{7}{3}\binom{3}{2}=  126

    Lastly the probablility is the quotient of the S_0+S_1 and S

    126/252=.5

    Which is the given answer. Could someone be kind enough to double check this for me?
    Last edited by bkbowser; January 28th 2014 at 05:19 PM. Reason: errant punctuation in title.
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  2. #2
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    Re: Odds of passing a test.

    Quote Originally Posted by bkbowser View Post
    An instructor gives her class a set of 10 problems with the information that the final exam will consist of a random selection of 5 of them. If a student has figured out how to do 7 of the problems, what is the probability that he or she will answer correctly (a) all 5 problems?
    Why would the answer is as simple as \frac{\dbinom{7}{5}}{\dbinom{10}{5}}~?

    Does that give the same answer? Why or why not?
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    Re: Odds of passing a test.

    Quote Originally Posted by Plato View Post
    Why would the answer is as simple as \frac{\dbinom{7}{5}}{\dbinom{10}{5}}~

    Does that give the same answer? Why or why not?

    It literally gives the same answer, but I can't construct a good argument around that binomial quotient.


    I had just noticed, during the process of making my solution, that that particular fraction spit out the same number and I was curious to see if I missed something or not.
    Last edited by bkbowser; January 29th 2014 at 08:17 AM. Reason: something goofy with laTex.
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