1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

    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).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook