2.6
a) P(sum=1) = 0. There are no combinations from rolling 2 dice that will give a sum of 0.
b) P(sum=5) = 4/36. Since there are 4 ways to roll dice to sum to 5 (1+4 or 2+3) and each die has a 1/6 probability of rolling one of those numbers (P(d1=1)=1/6 * P(d2=4)=1/6 = 1/36), the total probability will be 1/36 + 1/36 + 1/36 + 1/36.
c) P(sum=12) = 1/36. The only way to get 12 is with both dice being 6, so the probability will be 1/6 * 1/6 = 1/36.

2.8
a) Living below the poverty line and speaking a foreign language at home are not disjoint because there are people who are both below the poverty line and speak a foreign language.
b)

library(VennDiagram)
## Warning: package 'VennDiagram' was built under R version 3.3.3
## Loading required package: grid
## Loading required package: futile.logger
## Warning: package 'futile.logger' was built under R version 3.3.3
draw.pairwise.venn(.165, .104, .042, category = c("Speak a language other than English", "Live below the poverty line"), lty = rep("blank", 
    2), fill = c("light blue", "red"), alpha = rep(0.5, 2), cat.pos = c(0, 
    180), cat.dist = rep(0.025, 2), scaled = FALSE)

## (polygon[GRID.polygon.1], polygon[GRID.polygon.2], polygon[GRID.polygon.3], polygon[GRID.polygon.4], text[GRID.text.5], text[GRID.text.6], text[GRID.text.7], text[GRID.text.8], text[GRID.text.9])

c)
P(below poverty and only english) = P(below poverty) - P(below poverty and other than English)
P(below poverty and only english) = 14.6 - 4.2
P(below poverty and only english) = 10.4%

d)
P(below poverty or other than English) = P(below poverty) + P(other than English) - P(below poverty and other than English)
P(below poverty or other than English) = 14.6 + 20.7 - 4.2
P(below poverty or other than English) = 31.1%

e)
P(above poverty and only English) = 1-P(below poverty or other than English)
P(above poverty and only English) = 1 - .311
P(above poverty and only English) = 68.9%

f) P(below poverty and other than English) = P(below poverty) * P(other than enlgish)
31.1% > 14.6% * 20.7%
The event that someone lives below the poverty line is not independent of the event that the person speaks a foreign language at home

2.20
a) P(m_blue or f_blue) = P(m_blue)+P(f_blue) - P(m_blue and f_blue)
P(m_blue or f_blue) = 114/204 + 108/204 - 78/204
P(m_blue or f_blue) = 144/204 = 70.5%

b) (P f_blue | m_blue) = P(f_blue and m_blue) / P(m_blue)
(P f_blue | m_blue) = 78/204 / 114/204 = 68.4%

c) (P f_blue | m_brown) = P(f_blue and m_brown) / P(m_brown)
(P f_blue | m_brown) = 19/204 / 54/204 = 35.2%

(P f_blue | m_green) = P(f_blue and m_green) / P(m_green)
(P f_blue | m_green) = 11/204 / 36/204 = 30.6%%

d) The eye colors of male respondents and their partners are not independent since we see from using blue as an example the probability of having both blue is greater than having different colors.

2.30
a) P(hardcover then paperback ficiton) = 28/95 * 59/94 = 18.5%

b) P(fiction then hardcover no replacement) = 72/95 * 28/94 = 22.6%

c) P(fiction then hardcover replacement) = 72/95 * 28/95 = 22.3%

d) The answers to b and c are similar because there are a lot of can happen so having or not having the replacement of 1 book would not change the results by much.

2.38
a)

#Array of the fees for 0, 1, or 2 bags ($60 for 2 bags: 25 for first 35 for second)   
fees <- c(0,25,60)

#Array of the percent of passengers who check 0, 1 , or 2 bags
passengers <- c(.54,.34,.12)

E <- sum(fees*passengers)
E
## [1] 15.7
Vx <- round(sum((fees^2)*passengers) - E^2, 1)

SDx <- round((Vx)^(1/2),1)
SDx
## [1] 19.9

Expected value (avg rev per passenger) is 15.7 and the standard deviation is 19.9

b)

#Array of the fees for 0, 1, or 2 bags ($60 for 2 bags: 25 for first 35 for second)   
fees <- c(0,25,60)

#Array of the percent of passengers who check 0, 1 , or 2 bags
passengers <- c(.54,.34,.12)

E <- sum(fees*passengers)
E120 <- E*120 #multiply for 120 passengers
E120
## [1] 1884

They should expect a revenue of $1884 from 120 passengers with a standard deviation of 19.9

2.44
a)

income <- c("1 to 9,999","10,000 to 14,999","15,000 to 24,999","25,000 to 34,999","35,000 to 49,999","50,000 to 64,999","65,000 to 74,999","75,000 to 99,999","100,000 or more")

total <- c(2.2,4.7,15.8,18.3,21.2,13.9,5.8,8.4,9.7)

freq <- data.frame(income,total)
freq
##             income total
## 1       1 to 9,999   2.2
## 2 10,000 to 14,999   4.7
## 3 15,000 to 24,999  15.8
## 4 25,000 to 34,999  18.3
## 5 35,000 to 49,999  21.2
## 6 50,000 to 64,999  13.9
## 7 65,000 to 74,999   5.8
## 8 75,000 to 99,999   8.4
## 9  100,000 or more   9.7
barplot(freq$total, names.arg=income)

The distribution has a mean around 25,000 and 49,999 and peaks in the 35,000 to 49,999 range before decreasing again.

b)
P(< 50,000) = 2.2 + 4.7 + 15.8 + 18.3 + 21.2 = 62.2%

c)
P(female and <50k) = .622 * .41 = 25.5%

*d)**
The assumption in c would not be correct. With 41% of the population (96,420,486) being women we would have 39,532,399. From that population of women we would have 28,384,262 (71.8%) women making less than $50k. Using these metrics we get a probability of 29.4% of women making less than $50K (28,384,262/96,420,486).