Spelling errors in a text are either nonword errors or word

Spelling errors in a text are either \"nonword errors\" or \"word errors.\" A nonword error produces a string of letters that is not a word, as when \"the\" is typed as \"teh.\" Word errors produce the wrong word, as when \"loose\" is typed as \"lose.\" Nonword errors make up 30% of all errors. A human proofreader will catch 90% of nonword errors and 61% of word errors. Of all errors that the proofreader catches, what percent are nonword errors? % (Round your answer to the nearest whole answer.)

Solution

This is a binomial probability, the general equation is below. p(X = k) = (n choose k)p^k*(1-p)^(n-k) Note that getting 7 incorrect is the same as getting 3 correct. Let k = 3, n = 10, and p = .7, therefore: p(X = 3) = (10 choose 3)(.7)^3*(1-.7)^(10-3) p(X = 3) = 120(0.343)(0.0002187) p(X = 3) = 0.009001692 Probability of missing 6 or more is equal to getting 4 or less correct. Therefore p(X<=4) = p(X=0) + p(X=1) + p(X=2) + p(X=3) + p(X=4) p(X = 4) = (10 choose 4)(.7)^4*(1-.7)^(10-4) = 0.036756909 p(X = 3) = (10 choose 3)(.7)^3*(1-.7)^(10-3) = 0.009001692 p(X = 2) = (10 choose 2)(.7)^2*(1-.7)^(10-2) = 0.0014467005 p(X = 1) = (10 choose 1)(.7)^1*(1-.7)^(10-1) = 0.000137781 p(X = 0) = (10 choose 0)(.7)^0*(1-.7)^(10-0) = 0.0000059049 Add these all together and you get 0.0473489874