Summary: HOMEWORK SOLUTIONS FOR MATH 5651
HOMEWORK FOR WEEK 7 OF FALL 2011
Assignment from fourth edition of DeGroot and Schervish:
§3.6:2,4,8,10; §3.7:1,4,7; §3.8:2,3,8; §3.9:2,4,6,8
Sometimes I write P instead of Pr because it's shorter.
Problem 2 of §3.6
(Probability to be a junior)= (0.04 + 0.20 + 0.90)
(sum the junior row).
(Probability to be a junior and never visited museum)= 0.04.
(Probability to have never visited a museum given that one is a junior)
= (0.04)/(0.04 + 0.20 + 0.90) = .1212
We don't have data on students who have visited three or more times.
Best we can do is to calculate the probability that the student is a senior
given that she has visited a museum more than once.
(Probability that a student has visited more than once)= 0.04+0.04+0.09+0.10.
(sum the more-than-once column)
(Probability that a senior has visited more than once)= 0.10.
(Probability to be a senior given having visited a museum more than once)