Question

Typing errors in a text are either nonword errors (as when "the" is typed as "teh") or word errors that result in a real but incorrect word. Spell-checking software will catch nonword errors but not word errors. Human proofreaders catch 70% of word errors. You ask a fellow student to proofread an essay in which you have deliberately made 11 word errors.

(b)

Missing two or more out of 11 errors seems a poor performance. What is the probability that a proofreader who catches 70% of word errors misses exactly two out of 11? (Round your answer to four decimal places.)

What is the probability of missing two or more out of 11? (Round your answer to four decimal places.)

Answer 1

Let , X be the number of word errors.

Here , X has binomial distribution with parameter n=11 and p=0.70

Therefore , the probability mass function of X is ,

; x=0,1,2,..........,n and q=1-p

= 0 ; otherwise

1) Now , P(X=2)=0.0005 ; From binomial probability table

Therefore , the probability that a proofreader who catches 70% of word errors misses exactly two out of 11 is 0.0005

2) Now ,

Therefore ,  the probability of missing two or more out of 11 is approximately 1.