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Probability of Same Score on Test

  1. Jun 20, 2005 #1
    "Laura and Steve are taking a 50-question test with all questions worth 2 points and no partial credit. Assume that Laura and Steve have the same capability of performing well on the tests, that all the problems are equally difficult for these students, and that their expected score on the test is 50. What is the probability that they will get the same score?"

    I'm kind of introducing myself to probability, but haven't studied it too much besides the little bit I did back in Precalculus. Does the binomial theorem have something to do with this?

    Thanks for your help.
     
  2. jcsd
  3. Jun 20, 2005 #2
    You have a 50 X 50 matrix of binomial probabilities. You want to sum up the diagonal.
     
  4. Jun 21, 2005 #3
    The problem is incomplete. That the expected score on the test is 50 doesn't mean that the chance of correctly answering any given question is exactly 1/2.
     
  5. Jun 21, 2005 #4
    It does say the problems are equally difficult.
     
  6. Jun 21, 2005 #5
    I think the assumption they want you to make is that each question is independent and has a random probability p of being answered correctly.
     
  7. Jun 21, 2005 #6
    The only way to get an expected score of 50, if all the probabilities are p, is if p = 1/2.
     
  8. Jun 21, 2005 #7
    Ah, right.
     
  9. Jun 21, 2005 #8
    This is a weird question. It seems that the only possible results of the questions are "right" and "wrong"... usually on questions like these I've seen something along the lines of "their are 5 possible answers for each question". Their is no way of establishing a probability statement with an infinite number of answers.
     
  10. Jun 21, 2005 #9
    No, it is as simple as Juvenal described. It said there is no partial credit for the questions so you either get 0 points or 2 points. If there are 5 possible answers for each question, 4 are wrong with combined probability of being chosen of 1/2 and 1 is right with probability of being chosen of 1/2. So it can be reduced to 2 outcomes: wrong or right.
     
  11. Jun 21, 2005 #10
    Right, but with 5 answers, you can say their is a 1 in 5 chance of getting the question right from guessing. With only one possible correct answer and a infinite number of choices, I just don't see how you can say the probability is 1/2.
     
  12. Jun 21, 2005 #11

    NateTG

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    There are two possibilities:
    The correct answer is chosen. (2 points, probability .5)
    The correct answer is not chosen. (0 points, probability .5)
     
  13. Jun 21, 2005 #12
    I understand that much, but just logically thinking about it that doesn't seem right. I'm not disagreeing with your statement. Just thinking it through it seems that their is much more of a random chance that the answer is wrong than it is right.
     
  14. Jun 21, 2005 #13
    Yes, if the questions have more than two answer choices each then the students are not selecting with equal probability to each choice. They select correct/incorrect with equal probability, not a/b/c/d/e with equal probability. The students have some knowledge of the test material and are not just guessing.
     
  15. Jun 21, 2005 #14
    There is a less calculation-intensive way to do this problem than summing the diagonal. Find P(Laura's correct answers - Steve's correct answers = 0). Laura's and Steve's correct answers are distributed binomially and independently. So where X, Y ~ B(50, .5), find Z = X - Y, and then find P(Z = 0).
     
    Last edited: Jun 21, 2005
  16. Jul 10, 2005 #15
    alexmcavay@gmail.com: I'm kind of introducing myself to probability, but haven't studied it too much besides the little bit I did back in Precalculus. Does the binomial theorem have something to do with this?

    Rather than argue over possible complexities, I make a start on the problem, taking it in the simplest sense. The probability of any answer being correct is p=1/2, and thus 1-p=q=1/2. What we seek is the middle term in the binominal expansion of (p+q)^50. Thus we want to evaluate: [tex](pq)^-25\frac{50!}{25!25!}[/tex]

    My computer program Pari has no trouble with this and the probability is approximately 11.3%, and for both of them to have this score is about: 1.277%. (This is the greatest possibility, that both scores are 25.)
     
    Last edited: Jul 10, 2005
  17. Jul 10, 2005 #16

    HallsofIvy

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    The problem didn't say they were guessing. It said their expected score was 50. Since each of the 50 problems is worth 2 points, the expected number of questions answered correctly is 25, half of the 50 problems. Assuming the same probability for all questions ("all the problems are equally difficult for these students") the probability of answering any one correctly is 1/2.

    "With only one possible correct answer and a infinite number of choices, I just don't see how you can say the probability is 1/2."

    Consider this problem:
    1+ 1=
    (a) 1
    (b) 2
    (c) 3
    .
    .
    .

    What is the probability that you will answer correctly? :smile:
     
  18. Jul 10, 2005 #17
    Note: I am using the simplest approach to the problem. The student said of probabiliity: "I haven't studied it too much," so a simple answer seems best.

    To do this problem the easy hard way, assuming a 50% probability on each question, I just set up my FREE Pari program as such: sum(i=0,50, binomial(50,i)^2) =100,891,344,545,564,193,334,812,497,256. (Result seemed instant, around 100 octillion effortlessly). Then dividing by 2^100, make it a percent, and rounding off: 7.95892%, or almost 8% of the time two people will have the same score. If that is hard to believe on 50 questions, you must realize that the Standard Deviation is not much when the probability is 50%.. STD= sqrt(1/4*50)=3.54.

    Thus in roughly 70% of the cases the scores willl not vary as much as 21 to 29 answers correct. So that they cluster up near the average result of 25. (Or, getting all 50 correct is worth 1 part in 2^100, and is not likely at all.)
     
    Last edited: Jul 11, 2005
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