Human proof readers catch non-word errors 70% of the time. You ask a fellow student to proofread an essay in which you have deliberately made 10 word errors. Missing 4 or more out of 10 errors is poor performance.
What is the probability that a proof reader who catches 70% of word errors misses 4 or more out of 10?
What is the probability that a proof reader who catches 70% of word errors misses 4 or more out of 10?
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This is a binomial probability, the general equation is below.
p(X = k) = (n choose k)p^k*(1-p)^(n-k)
Probability of missing 4 or more is equal to getting 6 or less correct.
Therefore p(X<=6) = p(X=0) + p(X=1) + p(X=2) + p(X=3) + p(X=4) + p(X=5) + p(X=6)
p(X = 6) = (10 choose 6)(.7)^6*(1-.7)^(10-6) = 0.200120949
p(X = 5) = (10 choose 5)(.7)^5*(1-.7)^(10-5) = 0.102919345
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 0.350389281
From TutorTeddy
p(X = k) = (n choose k)p^k*(1-p)^(n-k)
Probability of missing 4 or more is equal to getting 6 or less correct.
Therefore p(X<=6) = p(X=0) + p(X=1) + p(X=2) + p(X=3) + p(X=4) + p(X=5) + p(X=6)
p(X = 6) = (10 choose 6)(.7)^6*(1-.7)^(10-6) = 0.200120949
p(X = 5) = (10 choose 5)(.7)^5*(1-.7)^(10-5) = 0.102919345
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 0.350389281
From TutorTeddy