More Exercises in Probability ANSWERS

for the Questions at http://student.ccbcmd.edu/elmo/math141s/practice/and_or2.htm

1. P(purple) = 0.000
2. P(pink & yellow) = 0.0451
3. P(black or even) = 0.692
4. P(both had corrected food-safety violations) = 0.047
5. P(at least one had major problems) = 99%
6. P(all smoked) = 6.25%
7. P(at least one goes to the movies at least once a month) = .971
8. Are the two events mutually exclusive?

Two ways of solving Problem 5: According to the Consumer Reports website, Consumer Union did a nationwide survey of owners of manufactured (mobile) homes. The report determined that 6 out of 10 (or 60%) of the people had major problems with their homes. If 5 manufactured home owners are randomly selected, what is the probability that at least one of them had major problems with their homes?

Solution by Addition: let '&' signify and an '|' or

• P = person 1 had major problems
• Q = person 2 had major problems
• R = person 3 had major problems
• S = person 4 had major problems
• T = person 5 had major problems

2. p(P|Q) = p(P) + p(Q) - p(P&Q) = .6 + .6 -(.6*.6) = .84
3. p(P|Q|R) = p(P|Q) + p(R) - p((P|Q) & R) = .84 + .6 - (.84*.6) = .936
4. p(P|Q|R|S) = p(P|Q|R) + p(S) - p(P|Q|R) & S) = .936 + .6 - (.936*.6) = .975
5. p(P|Q|R|S|T) = p(P|Q|R|S) + p(T) - p(P|Q|R|S & T) = .975 + .6 - (.975*.6) = .99

Solution by Multiplication plus Negation

1. p(P|Q|R|S|T) = 1 - p(not-(P|Q|R|S|T)) {negation theorem}
2. p(not-(P|Q|R|S|T)) = p(not-P & not-Q & not-R & not-S & not-T) {equivalently}
3. p(not-P & not-Q & not-R & not-S & not-T) = (.4*.4*.4*.4*.4) = .01 {applying the multiplication theorem}
4. p(P|Q|R|S|T) = .99 {applying the negation theorem}