Homework #4 -- Suggested Answers

ISBE 3.15, 3.20, 3.21, 3.22, 3.25, 3.26, 3.27

3.15

1. 7.2%
2. The conditional probability of randomly selecting an unemployed worker given the person is a male = 3.9%/55.8% = 6.99%
3. The conditional probability of randomly selecting an unemployed worker given the person is a female = 3.3%/44.2% = 7.47%

3.20

1. Pr(F|E)= Pr(total is 7 | first is 5)=1/6=16.7%, i.e. the probability that the second roll is a 2. Pr(F)=6/36 (6 of the 36 possible rolls yield a total of 7), hence Pr(F|E) = Pr(F).
2. Pr(G|E)= Pr(total is 10 | first is 5)=1/6=16.7%, i.e. the probability that the second roll is a 5. Pr(G)=3/36 (3 of the 36 possible rolls yield a total of 10), hence Pr(G|E) does not equal Pr(G).
3. When betting on a ten from a roll of two dice, it does change the odds in your favor if you know the first die is a 5. However, when betting on a total of 7, a peek doesn't help.

3.21

1. 7.8% + 18.2% = 26%
2. 7.8%/(7.8% + 22.2%) = 26% = (7.8/30)
3. Yes, if Pr(F|E) = Pr(F) (26%=26%), then being in favor of the legalization of marijuana is statistically independent of living in the East.

3.22

a. i. Yes, whenever F is independent of E, then E must be independent of F.

b. i. Pr(E|F) = 30%, Pr(E) = 30%

ii. Pr(F bar and E) = 22.2%, Pr(E)= 30%, thus Pr(F bar |E) = 22.2%/30% = 74% and equals Pr(F bar) = 22.2% + 51.8% or 74%.

3.25) Pr(cancer | + reaction) = Pr(+ reaction | cancer)*Pr(cancer)/Pr(+reaction)

3.26) Pr(rain | barometer says rain) = ((0.4 * 0.9)/(0.4*0.9 + 0.6*0.3)) = 0.667

3.27)

1. Pr(poor shipment) = 300/(300+900) = 1/4 = 0.25
2. Pr(poor shipment | draw defective saw) = ((0.25*0.4)/(0.25*0.4 + 0.75*0.1)) = 0.57