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Binomial function

  1. Jan 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Midvale School for the Gifted has two types of students: Guessers and Swots. All
    Midvale tests consist of sets of questions with yes/no answers. Guessers will simply
    answer yes or no to each question as the mood takes them, so they have probability
    0.5 of getting each question correct. Swots, on the other hand, will certainly get
    every question correct with probability 1. Dr Ramble Fluster is the Chief Examiner
    at Midvale. She worries about her test results. Clearly, all Swots will get 100% on
    every test. However, it is possible that a Guesser might get 100% just by chance.
    Ramble's job is to decide how many questions to set in each test to be fairly certain
    that she will be able to give the correct grade (G or S) to each student.

    (b) What is the probability that a Guesser scores 100% on a test with n questions?


    2. Relevant equations
    Binomial function


    3. The attempt at a solution
    X~Bin(n,0.5)

    P(X=x)= nCx * p^x * (1-p)^(n-x)

    P(X=100) = nC100 * 0.5^100 * 0.5^(n-100)

    here is where I got stuck. Since the question is asking us to solve the probability equation (P(X=x)), but since n is an unknown I don't know how to go further into solving this equation, because (I presume) there will be 2 unknown parameters: P and n.

    can someone point me where I get wrong, or give me an idea on how to solve this?
    thanks
     
  2. jcsd
  3. Jan 22, 2012 #2

    Dick

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    nC100?? Think about what that means. You aren't choosing 100 percentage points from n percentage points. Scoring 100% means you got all n questions right. Shouldn't the probability involve nCn?
     
  4. Jan 22, 2012 #3
    ar! ok, I've just reattempt this question from what I understand in your suggestion. Which doesn't seems to be correct though, need more help thanks.

    reattempts are as follows:

    fx(x)= p(X=x) = nCx * p^x * (1-p)^n-x

    let x = n

    p(X=n) = nCn * p ^ n * (1-p)^n-n
    = 1 * 0.5^n * 0.5^0
    = 0.5^n
     
  5. Jan 22, 2012 #4

    Dick

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    That seems ok to me. E.g. if there are two questions P=(1/4)=(0.5)^2. You have to get the first question right AND the second question right - both with probability 1/2.
     
  6. Jan 22, 2012 #5
    ah! looks light you are right. because I just read the next part of the question "Ramble wishes to select the number of questions, n, so that there is less than 0.5% chance that a Guesser gets full marks. What is the lowest value of n that will guarantee this?"

    so I will just substitute in the numbers and found out what n is.

    thanks a lot for your help!
     
  7. Jan 22, 2012 #6

    Dick

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    You're welcome. You could solve it systematically using a log. But there's nothing wrong with substituting either.
     
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