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Probablility equations

  1. Jul 27, 2007 #1
    1. The problem statement, all variables and given/known data

    A ten question multiple choice exam is given, and each question has 5 possible answers. Pascal Gonyo takes this exam and guesses at every question. Find the probability that he will get at least 9 questions correct. Without using the binomial probability formula, determine the probability that he gets exactly two questions correct.

    2. Relevant equations
    P(at least 9) + P(less than 9) = 1

    3. The attempt at a solution

    P = 1-P(8)
    I'm stuck on this. I found the probability of at least one question correct using this formula but I can't get it to work for 9.
  2. jcsd
  3. Jul 27, 2007 #2
    Use the binomial formula for the first part of the question.
  4. Jul 27, 2007 #3


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    Homework Helper

    The probability he will get exactly two questions correct is pretty easy. What is it? As is the probability he will get at least 9 correct is the sum of the probabilities of getting getting exactly 9 correct plus the probability of getting exactly 10 correct. What are they? I'm not sure I see a way to do these without using at least a vestige of the binomial theorem. But someone may prove me wrong tomorrow.
  5. Jul 28, 2007 #4
    The binomial formula will give me exactly 9 right? how do I get at least 9?
  6. Jul 28, 2007 #5
    Add the prob of getting exactly 9 and the prob of getting exactly 10.
  7. Jul 28, 2007 #6
    Thanks, can you tell me how to find the probability of getting 2 correct without the binomial formula?
  8. Jul 28, 2007 #7
    It's a dumb question! Why on earth should they tell you not to use a particular theorem?
  9. Jul 28, 2007 #8
    OK, so how many distinct possibilities are there for scoring?

    e.g. for 3 Q's there's 8 possibilities


    For 10 Q's how many distinct possibilities are there?

    Oops- there's 5 possible answers per Q. Well- can you work it out for that case?
  10. Jul 28, 2007 #9
    I am quite sure the OP is just looking for some clarity, dont be that harsh.o:)
  11. Jul 28, 2007 #10


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    Perhaps because they want people to think rather than just apply formulas? It wouldn't make sense to work it out from first principles for, say, 4 or 5 out of 10, but for "exactly 2" it shouldn't be too much to expect.

    What is the probability of "CCIIIIIIII" where C= correct and I= incorrect?
    Can you show that the probability of 2 correct and 8 incorrect is exactly the same no matter what order?
    How many different orders are there- in how many ways can you write 2 Cs and 8 Is?
  12. Jul 28, 2007 #11
    Good grief!!

    Threetheoreem: I was criticizing the Q, not the poster. The poster presumably didn't compose the question.

    HOI: I said it's a dumb question because I don't like the idea of restricting how a student solves a problem. Give me the Q- but don't tell me how to find the A. It also takes intelligence to see if the binomial theorem is applicable in this case.
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