3-4 (p. 78)
We can obtain these probabilities using the tree illustrated in Figure 3-3 (p. 75), or we can obtain them as outlined below. Remember these are families of three children and that we are using the probabilities:
P(G) = P(girl) = 0.48
P(B) = P(boy) = 0.52
a.P(exactly 2 girls)=(0.48)(0.48)(0.52) + (0.48)(0.52)(0.48) + (0.52)(0.48)(0.48) = 0.36 (approx)

b.P(at least 2 girls) = P(exactly 2 girls or exactly 3 girls) = P(exactly 2 girls) + (0.48)(0.48)(0.48) = 0.47 (approx)

c.P(at least one child of each sex) = 1 - P(all 3 boys) - P(all 3 girls) = 1 - (0.48)(0.48)(0.48) - (0.52)(0.52)(0.52) = 0.75 (approx)

d.P(the middle child opposite in sex to the other 2) = P(girl, boy, girl) + P(boy, girl, boy) = (0.48)(0.52)(0.48) + (0.52)(0.48)(0.52) = 0.25 (approx)

3-14 (p. 85)
a.P(no sales) = (0.80)^12 = 0.07 (approx)
b.P(at least 1 sale) = 1 - P(no sales) = 0.93
c.If he sells for 200 days of the year, he can be expected to make at least one sale on approximately (0.93)(200) = 186 of those days.

3-18 (p. 89)
a.P(2 are good) = P(1st good)P(2nd good | 1st good) =(6/10)(5/9) = 30/90 = 1/3
b.P(next 3 are good | first 2 are good) = (4/8)(3/7)(2/6) = 1/14 = 0.07 (approx)
c.P(5 are good) = (6/10)(5/9)(4/8)(3/7)(2/6) = 1/42 = 0.02 (approx)

3-20 (p. 89)
E: event that first die is 5
F: event that total of dice is 7
G: event that total of dice is 10
P(E)=1/6, P(F)=6/36=1/6, P(G)=3/36=1/12
a.P(F|E) = P(F and E)/P(E) = (1/36)/(1/6) = 1/6
b.P(G|E) = P(G and E)/P(E) = (1/36)/(1/6) = 1/6
c.If you're going to place a bet on whether the sum is seven, knowing that one of the dice shows a five doesn't help. That's because any of the outcomes (1-6) could be used just as easily to come up with a sum of seven. When you're betting on whether the sum is ten, it would be good to know that one of the dice shows five. This is because not all of the outcomes are conducive to obtaining a sum of ten. For instance, if you looked at the outcome of the first die, and it was one, two, or three, you would want to retract your bet on the sum of ten, because there would then be no way to obtain a sum of ten (no matter the value of the second die).

3-26 (p. 97)
Designate the events as follows:
W: event that the barometer "warns of" (predicts) rain
R: event that it rains
We want the probability that it actually rains, given that the barometer predicts rain.

P(R|W) = P(W|R)P(R) / ( P(W|R)P(R) + P(W|not R)P(not R) )
P(R|W) = (0.90)(0.40) / ( (0.90)(0.40) + (0.30)(0.60) )
P(R|W) = 0.36 / 0.54 = 2/3

3-34 (p. 103)
a.P(club) = 1/4; odds for a club: 1/3
b.P(black card) = 1/2; odds for a black card: 1/1
c.P(ace) = 1/13; odds for an ace 1/12
d.P(ace, king, queen, or jack) = 4/13; odds for ace, king, queen, or jack: 4/9
e.P(same denomination as next card) = 3/51; odds for this: 3/48
More explanation:
P(same denomination as next card) = P(get an ace 1st)P(get an ace 2nd|get an ace 1st) + P(get a 2 1st)P(get a 2 2nd|get a 2 1st) + ... + P(get a king 1st)P(get a king 2nd|get a king 1st) --> for all 13 denominations
P(same denomination as next card) = (4/52)(3/51) + (4/52)(3/51) + ... + (4/52)(3/51) = 13(4/52)(3/51) = 3/51
f.P(higher denomination than next card) = P(not lower ranking card on 1st try) P(lower ranking card on 2nd try| not lower ranking card on 1st try)
P(higher denomination than next card) = P(get a 2 1st)(P get 1 2nd|get a 2 1st) + P(get a 3 1st)P(get a 1 or 2 2nd|get a 3 1st) + ... + P(get a king 1st)P(get anything lower than a king 2nd|get a king 1st)
P(higher denomination than next card)= (4/52)[(4/51) + (8/51) + (12/51) + (16/51) + (20/51) + (24/51) + (28/51) + (32/51) + (36/51) + (40/51) + (44/51) + (48/51)] = (4/52)(312/52) = 0.47 (approx)
odds for this: 0.88 (approx)