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Probability of true/false exam

  1. Sep 14, 2013 #1
    1. The problem statement, all variables and given/known data
    Information in condensed form:

    => Student takes 20 questions true or false exam.
    => He knows answer to 10 questions.
    => He guesses on 10 questions.
    => Professor randomly picks two questions out of 20, and finds student answered them both correctly.

    Task: Find probability student knew answer to at least one of the two questions.

    2. Relevant equations
    Bayes' Rule and conditional probability definition.


    3. The attempt at a solution


    My work. I found P(K|C) = 2/3 via Bayes' rule, where K: student knows answer, C: he got it correct. However I'm not sure how to tackle "at least 1" part of problem. My thoughts were pick c1, c2 element of C and find P(K|(c1 or c2)), but this seems like a difficult calculation.
     
  2. jcsd
  3. Sep 14, 2013 #2
    Hello,

    Could you please explain how you found P(K|C) ?
     
  4. Sep 14, 2013 #3
    Sure.

    I defined events as follows
    K: Student knows answer
    G: Student guesses answer
    C: Student gets question correct

    P(K) = P(G) = 1/2
    Since student knows 1/2 questions and guesses on rest.

    P(C|K) = 1, since if student knows answer he is certain to get it correct
    P(C|G) = 1/2, half-chance guessing true or false

    P(C) = P(C|K) * P(K) + P(C|G) * P(G) = 3/4

    P(K|C) = P(C|K) * P(K) / P(C) = (1 * 1/2) / (3/4) = 2/3
     
  5. Sep 14, 2013 #4

    vela

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    Isn't it just a binomial distribution problem now with N=2 and p=2/3?
     
  6. Sep 14, 2013 #5
    I thought the binomial distribution only applies when the trials are independent. Wouldn't the probability of getting the second question correctly depend on whether or not the first question was answered correctly or not?
     
  7. Sep 14, 2013 #6

    vela

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    Yeah, I think you're right.
     
  8. Sep 14, 2013 #7

    D H

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    That isn't correct, for a couple of reasons. One reason is that the student only answered 20 questions. You have to account for the fact that the universe is finite. The other is that the instructor randomly picks two questions out of the twenty to examine. You need to account for this as well.
     
  9. Sep 14, 2013 #8
    I based my formulation on the basis for one particular question out of 20, and hoped I could apply it to two questions. It may be easier to define events for two questions, but I wasn't sure on how to do so in a manner that would allow me to calculate the probabilities.

    For instance, if I define an event A such that the student knew the answer to both questions. I'm not sure how to calculate this probability because for one question it would 3/4, but for two questions this probability would decrease.
     
  10. Sep 14, 2013 #9

    D H

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    Hint: Look at what the instructor drew. He might have picked two questions for which the student knew the answer, two questions where the student was clueless and guessed, or one of each. These are mutually exclusive.

    What are the probabilities of these outcomes?
    For each of those outcomes, what is the probability that the student got both answers correct?
     
  11. Sep 14, 2013 #10
    Thanks a lot, I think I figured it out.

    Let Case 1: Student knew answer to both questions
    P(Case 1) = (10 C 2) / (20 C 2) = 9/38
    P(both correct | case 1) = 1

    Let Case 2: Student guessed both questions
    P(Case 2) = (10 C 2) / (20 C 2) = 9/38
    P(both correct| case 2) = (1/2)^2 = 1/4

    Let Case 3: Student knew one question and guessed on other
    P(Case 3) = 1 - P(Case 1) - P(Case 2) = 20/38 = 10/19
    OR P(Case 3) = [(10 C 1)(10 C 1)] / (20 C 2) = 10/19
    P(both correct| case 3) = 1 * 1/2 = 1/2

    we want:
    P((case 1 or case 3) | both correct) =
    [P(both correct | (case 1 or case 3)) * P(case 1 or case 3)] / (P(both correct))

    Since the cases are mutually exclusive P(case 1 or case 3) = P(case 1) + P(case 3) = 29/38

    P(both correct | (case 1 or case 3)) =
    P(both correct and (case 1 or case 3)) * P(case 1 or case 3) =
    P((both correct and case 1) or (both correct and case 3)) / P(case 1 or case 3) =
    [P(both correct and case 1) + P(both correct and case 3)] / P(case 1 or case 3) =
    [P(both correct|case 1) * P(case 1) + P(both correct|case 3) * P(case 3)] / P(case 1 or case 3)
    = [1 * 9/38 + 1/2 * 20/38] / [29/38] = 19/29

    P(both correct) = P(both correct|case 1) * P(case 1) + P(both correct|case 2) * P(case 2) + P(both correct| case 3) * P(case 3) = (1 * 9/38) + (1/4 * 9/38) + (1/2 * 20/38) = 85/152

    so...back to what we wanted
    P((case 1 or case 3) | both correct) =
    [P(both correct | (case 1 or case 3)) * P(case 1 or case 3)] / (P(both correct)) =
    (19/29 * 29/38) / (85/152) = 76/85.

    About 89.4%
     
    Last edited: Sep 14, 2013
  12. Sep 14, 2013 #11

    D H

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    Very good.
     
  13. Sep 14, 2013 #12

    Ray Vickson

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    Well, I get an answer of 76/85 ≈ 0.894.

    Here is what I did: for k guessed correct, the population consists of three classes: (1) 10 known & correct; (2) k guessed and correct; (3) 10-k guessed wrong. The probability of k guessed correct is the binomial probability Pk = C(10,k)/2^10, where C(a,b) denotes the binomial coefficient "a choose b".

    Denote by P(u,v,w) the probability that the prof. chooses u from class 1, v from class 2 and w from class 3. So, conditioned on k guessed OK, the probability that the prof chooses 2 correct is P(2,0,0)+P(1,1,0)+P(0,2,0). These are 3-class hypergeometric probabilities: that is, P(u,v,w) = C(10,u)*C(k,v)*C(10-k,w)/C(20/2) [all conditioned on k guessed OK]. Thus, we have P(2,0,0) = (10/20)(9/19), P(1,1,0) = 2(10/20)(k/19), and P(0,2,0) = (k/20)((k-1)19)= k(k-1)/(20*19).

    The probability that the prof chooses 2 correct and the student knows at least one is P(2,0,0) + P(1,1,0).

    So: P{2 correct} = sum_{k=0}^10 C(10/k)/2^10 * [P(2,0,0)+P(1,1,0)+P(0,2,0)] = 85/152.
    P{2 correct and >=1 known} = sum_{k=0}^10 C(10,k)/2^10 *[P(2,0,0)+P(1,1,0)] = 1/2.
    Thus, the conditional probability P{>=1 known|2 OK} = (1/2)/(85/152) = 76/85.
     
  14. Sep 14, 2013 #13
    Thanks for the catch. I got that answer first as well, but thought I made an arithmetic mistake, so I got the new one, but looking back 76/85 was indeed correct. Seems the methods are similar, but your terminology and formulation is better, because you can generalize this to variations of the problem. Thanks for the help.
     
    Last edited: Sep 14, 2013
  15. Sep 14, 2013 #14

    D H

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    That was the answer posted when I wrote "very good". I would not have written that had Gridvvk posted an answer of 58/85. Apparently Gridvvk wavered on the result for a while.
     
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