STAT/MATH 394B

Solutions, homework 4.

2.3.1. (a) .8 x .35 + .45 x .4 + .55 x .25 = .598

(b) P(Dem | good job) = P(Dem and good job)/ P( good job) =(.8 x .35)/ .598=.469

2.3.2

2.3.4 P(p=3/4|SSS)=P(SSS|p=3/4) x P(p=3/4) / P(SSS)

We have P(SSS)=P(SSS|p=3/4)P(p=3/4) + P(SSS|p=1/2)P(p=1/2) + P(SSS|p=1/4)P(p=1/4)

=(3/4)3 x 0.3 + (1/2)3 x 0.4 + (1.4)3 x 0.3 = .181

whence P(p=3/4|SSS) =.127/.181 = .698

2.3.7 (a)

Hence P(HIV|+)=.000294//(.00294+.004999)=.0555

(b) Only the numbers in the circles change, so the answer is then .1 x .98 / (.1 x .98 + .9 x .005) = .956

2.3.8

The conditional probability is then 3/11 / (3/11 + 5/22) = 6/11. The reverse conditional probability can be read off the diagram, and is the same.

2.3.11 Let p be the probability of heads, so p= 1/2 or 1 depending on the coin. Now

P(p=1|HH) = P(HH|p=1)P(p=1)/{P(HH|p=1) P(p=1) + P(HH|p=1/2) P(p=1/2)} =

1 x .1 / (1 x .1 + (1/2)2 x .9) = 22 / (22 + 9) = 4/13

For three heads only the power of 1/2 changes, so for n=3 we get 23/(23+9)=8/17, and

generally 2n/(2n+9). In other words, we become more and more sure that it is the two-headed coin we are seeing.