Practice Exam 2 21. Spelling errors in a text consist of two disjoint types: non

Question

Practice Exam 2 21. Spelling errors in a text consist of two disjoint types: non-word errors, and word errors. 75%of spelling errors are non-word errors probability that a randomly selected spelling error is a word error that will be caught by a spelling errors are typically caught by a human proofreader. If 225% of the human proofreader? Asurvey of college students indicates that 38.5% of them regularly listen to country music. and that 125% of them regularly listen to both country music and R&B; i, a randomly selected college student regularly listens to country music, what is the probability that the college student also regularly listens to R&cB;? For questions "23 and "24, show your work. I expressions in place. 24, show your work. If a counting formula is used, show its implementation with correct numeric 23. 7equally qualified students apply to a scholarship fund, but only 5 scholarships, each of different value, can be given. In how many ways can the 5 winners be selected? "24. 24. In purchasing a home theater system, you have 4 choices for speaker sets, another 4 choices for receivers, and 3 choices for TV. How many different home theater systems carn you construct consisting of one speaker set, one receiver, and one TV? For questions *25 through 27, use the following information The mayor of a small cty s rsquired to appoint from a 7-member appropriations commite a 3-person panel to review applications for tax exemptions requested by private citizens. For questions "25 and "26, show your work. If a counting formula is used, show its implementation with correct numeric values in place. *25. 25. In how many ways can this panel be appointed? 26. 26. The appropriations committee has 4 members who are Democrats. In how many ways an all-Democrat 3-person panel be appointed?

Explanation / Answer

Question 21:

Here, we are given that:

P( human proofreader | spelling errors ) = 0.75
P( non word errors , human proofreader | spelling errors ) = 0.225

Now the probability that a randomly selected spelling error is a word error that will be caught by a human proofreader here is computed as:

P( word errors , human proofreader | spelling errors ) = P( human proofreader | spelling errors ) - P( non word errors , human proofreader | spelling errors )

P( word errors , human proofreader | spelling errors ) = 0.75 - 0.225 = 0.525

Therefore 0.525 is the required probability here.

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