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Probability Distributions

  1. Feb 23, 2008 #1
    On a multiple guess exam, there are 3 possible answers for each of the 5 questions. What is the probability that the student will get four or more correct answers just by guessing?

    Is this hypergeometric or binomial?
  2. jcsd
  3. Feb 23, 2008 #2


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    Since on each question a student can either choose the correct answer or not, it is binomial. Do you see what is the probability a student will choose the correct answer on a specific question just by guessing?
  4. Feb 23, 2008 #3
    I thought it was either this

    [tex]\frac{(^{5}_{4}) (^{10}_{1})}{(^{15}_{5})}[/tex] choose 4 from 5 correct answers, choose 1 from 10 wrong answers divided by choosing 5 answers from 15

    Or this (which I now know is probably the right answer)

    [tex](^{5}_{4})p^{4}q + (^{5}_{5})p^{5}[/tex] which is the probability of 4/5 right answers + probability of 5/5 right answers, where prob correct answer = 1/3
  5. Feb 24, 2008 #4


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    Yes, the second is the one you want (you are NOT choosing 4 or 5 correct answers from all 15- that would imply that you could choose 2 correct answer from one problem and none from another!). Now what number is [tex](^{5}_{4})p^{4}q + (^{5}_{5})p^{5}[/tex]?
  6. Feb 24, 2008 #5
    p = 1/3 and q = 2/3 => 0.14 unless I calculated wrong which is entirely possible.
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