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Probability; counting questions

  1. Dec 6, 2007 #1
    Let me prefix by stating that this is not really a homework problem, just something I am curious about. I posted it here because it is probably too easy to go into the "big boy" forums :smile:.

    Last night my wife said she needed a 23% on her final to maintain a grade of A in her course. I asked her if the test was multiple choice, she said yes and usually the questions were 4 choice answers. I asked how many questions were usually in the exam, she said about 80. I said, "no need to study, you can probably guess randomly and have a really good chance of getting a 23% or better on the exam." Then I decided to put my theory to the test.

    I put together a simulator to "randomly generate" an 80 question test and then "randomly answer" the test and score it. After running about 2 million trials, the simulator says about 64% of them score >= 23%. So, I have my answer...now...how to get there with math.

    I scaled the problem down to a 2 question test. I assume a lower probability to score > 23%, and it seems I am correct; my simulator says 44%. This I think I can do in math:

    I solved this by chosing a strategy, "answer only A" (assuming a random distribution of answers on the test). Given a two question test, I need to answer either one or both questions correctly to score > 23%. So, there are 4^2 different random answer keys, and of them 7 have one or more A's in the key. So 7/(4^2) = 7/16 = .437 or 44% that my simulator says. Excellent.

    So...can I extend this solution method to the 80 question test? Is there an easier way to do this? If I can extend it to the 80 question test, how do I do the counting?

    Anyone want to help enlighten an intrigued mind.... I wish I could remember my game theory class, lol. I bet somewhere in those notes I'd find how to solve this, alas it was just too long ago and my mind is dusty. Please bring on the Pledge :smile:.
  2. jcsd
  3. Dec 6, 2007 #2


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    In general, it'd be something like this:
    Suppose you have n questions (in your (wifes) case, n = 80). For a given percentage, you will need k or more questions answered correctly (for 23% on 80 questions that's 19 questions). Assume that, on a random guess, you have a probability p to answer any question correctly (if all of four answers are equally likely, you would have p = 0,25; but in practice I think it will be around 0,3 or 0,4 because there are usually one or two answers which are clearly not correct).

    Now the problem is what mathematicians call a binomial distribution (click link for a lot of theory), because the chance of success (answering any given question correctly) is independent. Basically the idea is as follows: first consider the chance to get exactly k right. The chance that I answer the first k of n questions correctly, is p^k, and then the chance that the rest is wrong is (1 - p)^(n - k), so the total chance is p^k (1 - p)^(n - k). Then we multiply this by a number (binomial coefficient, n over k) to account for the other orderings (e.g. first correct, second incorrect, then 3 correct, then 5 incorrect, etc. - in such a way that k out of n are right and the rest is wrong).
    Repeat this calculation for k + 1 correct, k + 2 correct, etc. all the way up to n out of n correctly answered, and sum all the probabilities.
    (This is the same as 1 - (sum of all probabilities for 1 up to k - 1 right answers). This will produce a cumulative binomial function (cumulative, because it adds all results for several possibilities) which can easily be evaluated on an advanced pocket calculator or numerical computer program.

    It's all a bit sketchy, and I don't really know your level of mathematics, so I'll leave it at this for now; if anything is unclear, please just ask.
  4. Dec 6, 2007 #3
    Ahhh...well, that is easier than my method. Better yet, it produces the same answer.

    With the easier problem (2 questions), I opted to compute using the inverse (1 - the probability that you get everything wrong). Obviously, for a 2 question problem that is the only way you can get less than the desired score (1 correct and you have a 50%). 1 - (.75 ^ 2) = 7/16. Splendid, same as my "method."

    However, just for my edification. for the 80 question version of "my method," I would be saying, "there are 4^80th ways to key an 80 question test, of those y have 19 or more "A" answers. Therefore the probability that you select one of those keys (and therefore score 23% or better) is y / (4^80)." However, I am still stuck on computing "y". It seems y would be:

    / f(i)
    i = 19

    where f(x) = the number of ways to chose x slots of 80 for "A" where the other 80 - x slots are "not A".

    Is f(x) also a binomial distribution? It seems to be a counting problem for combinations/permutations. Maybe that's what is hanging me up....
  5. Dec 6, 2007 #4
    f(x) can be written as the binomial coefficient [itex]\binom{80}{x}[/itex]. It is not a distribution (that is something completely different).
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