BAYES THEOREM -In a certain year, 87% of all Caucasians in the U.S., 73% of all

Question

BAYES THEOREM

-In a certain year, 87% of all Caucasians in the U.S., 73% of all African-Americans, 73% of all Hispanics, and 85% of residents not classified into one of these groups used the Internet for e-mail. At that time, the U.S. population was 68% Caucasian, 12% African-American, and 12% Hispanic. What percentage of U.S. residents who used the Internet for e-mail were Hispanic? (Round your answer to the nearest whole percent.)

-In a survey in a certain year of married couples with earnings, 91% of all husbands were employed. Of all employed husbands, 71% of their wives were also employed. Noting that either the husband or wife in a couple with earnings had to be employed, find the probability that the husband of an employed woman was also employed. (Round your answer to four decimal places.)

In a certain year, 87% of all Caucasians in the U.S., 73% of all African-Americans, 73% of all Hispanics, and 85% of residents not classified into one of these groups used the Internet for e-mail. At that time, the U.S. population was 68% Caucasian, 12% African-American, and 12% Hispanic, what percentage of U.S. residents who used the 17 Internet for e-mail were Hispanic? (Round your answer to the nearest whole percent.) Need Help?Read ItMaster t

Explanation / Answer

1. a)

Let I = Internet e-mail user C = Caucasian
H = Hispanic A = African_American
O= Other

P(I/C) = .87 (probability that one used e-mail given that he is Caucasian)
P(I/H) = .73
P(I/A) = .73
P(I/O) = .85

P(C)=.68
P(H)=.12
P(A)=.12
P(O)=.08

We want to find:
P(H/I) = P(H) P(I/H) / [ P(C) P(I/C) + P(H) P(I/H)+P(A) P(I/A)+P(O)P(I/O)]
P(H/I) = (.12)(.73) / [ (.68)(.87)+(.11)(.73)+(.12)(.73)+(.08)(0.85)
=0.1059
= 10.59 %

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.