Lets say a person decided to randomly fill in a scantron of 50 questions with 4 choices each. After submitting it to be graded, the result was 42% correct. How would we figure out the probability of this occurring completely by chance?

This problem actually came up while I was watching a show. I would assume that a probability test is to be used here. Observed is 21, while the expected is 12.5 questions. How would one go from here?


Assuming each the answer to each question is independently chosen, the answer should be the number of ways we can order 21 correct answers and 29 incorrect answers.

A correct answer has a probability of $1/4$ for an individual question, and $3/4$ for an incorrect answer.

There are ${50 \choose 21}$ ways to select the 21 correct answers out of 50 questions. So the final probability should be $$ {50 \choose 21} (1/4)^{21}(3/4)^{29} $$ This works out to about 0.3%

  • $\begingroup$ Oooh yes, that seems right. I guess I should have clarified in my question that I wanted to figure out the probability of getting 42% or better, which would just be the summation of your formula from 21 to 50. Any idea on if there is a more clean way of writing and calculating that though? $\endgroup$ – MaximumMath Mar 5 '14 at 18:48
  • $\begingroup$ @SeitokaiShino: as an approximation, adding another correct reduces the probability by about a factor $3$ as the combinatoric factor doesn't change much. The sum of $1+\frac 13+\frac 1{3^2}+\dots=\frac 32$, so the total will be about $\frac 32$ times the value above. $\endgroup$ – Ross Millikan Mar 5 '14 at 18:56

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