Chapter 2 - Probability

2.6, 2.8, 2.20, 2.30, 2.38, 2.44

2.6 Dice rolls. If you roll a pair of fair dice, what is the probability of

  1. getting a sum of 1? 0 because each die will at least equal 1, so the smallest possible sum is 2

  2. getting a sum of 5? To get a sum of 5, one die has to be a 2 and the other a 3, or one die has to be a 1 and the other a 4. The probability of the first die being a 2 and the second a 3 is (1/6)*(1/6). The probability of the first die being a 3 and the second a 2 is also (1/6)^2. Similarly, we have the same probabilities for a die of 1 and a die of 4. Adding these up, the probability of getting a sum of 5 is 4/36 = 1/9.

  3. getting a sum of 12? The only way to get a sum of 12 is for both dice to be 6. This is (1/6)*(1/6)=1/36

2.8 Poverty and language. The American Community Survey is an ongoing survey that provides data every year to give communities the current information they need to plan investments and services. The 2010 American Community Survey estimates that 14.6% of Americans live below the poverty line, 20.7% speak a language other than English (foreign language) at home, and 4.2% fall into both categories.

  1. Are living below the poverty line and speaking a foreign language at home disjoint? Living below the poverty line and speaking a foreign language at home are not disjoint as there are cases that fall under both categories.

  2. Draw a Venn diagram summarizing the variables and their associated probabilities. In this venn diagram, living below the poverty line is on the left with 10.4%, speaking a langauge other than English is on the right with 16.5%, and their intersection is in the middle with 4.2%.

  3. What percent of Americans live below the poverty line and only speak English at home? 10.4% of Americans live below the poverty line and only speak English at home.

  4. What percent of Americans live below the poverty line or speak a foreign language at home? 14.6% + 20.7% - 4.2% = 31.1% of Americans live below the poverty line or speak a foreign language at home.

  5. What percent of Americans live above the poverty line and only speak English at home? This would be the area outside of our venn diagram, which is 1 - (10.4 + 16.5 + 4.2) = 68.9

  6. Is the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home? Two events are independent if P(A|B) = P(A). Here, P(living below the poverty line | speaking a foreign language at home) = .042 / .207 = .203 != .146 = P(living below the poverty line). So, these events are not independent.

2.20 Assortative mating. Assortative mating is a nonrandom mating pattern where individuals with similar genotypes and/or phenotypes mate with one another more frequently than what would be expected under a random mating pattern. Researchers studying this topic collected data on eye colors of 204 Scandinavian men and their female partners. The table below summarizes the results. For simplicity, we only include heterosexual relationships in this exercise.

  1. What is the probability that a randomly chosen male respondent or his partner has blue eyes? The probability that a male respondent has blue eyes is P(A) = 114/204. The probability that his partner has blue eyes is P(B) = 108/204. P(A or B) = P(A) + P(B) - P(A and B) = 114/204 + 108/204 - 78/204 = 144/204 = 0.706.

  2. What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes? This would be 78/114 = 0.684.

  3. What is the probability that a randomly chosen male respondent with brown eyes has a partner with blue eyes? What about the probability of a randomly chosen male respondent with green eyes having a partner with blue eyes? P(partner blue eyes | male with brown eyes) = 19/54 = 0.352 P(partner blue eyes | male with green eyes) = 11/36 = 0.306

  4. Does it appear that the eye colors of male respondents and their partners are independent? Explain your reasoning. For male respondents with green eyes having a partner with blue eyes, P(partner blue eyes | male with green eyes) = (11/204) / (36/204) = 11/36 != 108/204 = P(partner blue eyes). So, the eye colors of male respondents and their partners are not independent.

2.30 Books on a bookshelf. The table below shows the distribution of books on a bookcase based on whether they are nonfiction or fiction and hardcover or paperback.

  1. Find the probability of drawing a hardcover book first then a paperback fiction book second when drawing without replacement.
(28/95) * (59/94)
## [1] 0.1849944
  1. Determine the probability of drawing a fiction book first and then a hardcover book second, when drawing without replacement.
(72/95) * (28/94)
## [1] 0.2257559
  1. Calculate the probability of the scenario in part (b), except this time complete the calculations under the scenario where the first book is placed back on the bookcase before randomly drawing the second book.
(72/95) * (28/95)
## [1] 0.2233795
  1. The final answers to parts (b) and (c) are very similar. Explain why this is the case.

These answers are very similar because the denominator does not change by much when we return the book to the bookcase and select from a bookshelf that has one more book than it had in b’s scenario.

2.38 Baggage fees. An airline charges the following baggage fees: $25 for the first bag and $35 for the second. Suppose 54% of passengers have no checked luggage, 34% have one piece of checked luggage and 12% have two pieces. We suppose a negligible portion of people check more than two bags.

  1. Build a probability model, compute the average revenue per passenger, and compute the corresponding standard deviation.

The average revenue per passenger is $15.7. The standard deviation is $19.25.

price <- c(0, 25, 60)
checking <- c(.54, .34, .12)
bags <- c(0, 1, 2)
t <- data.frame(price, checking, bags)
t$comp <- price * checking
avgRev <- sum(t$comp)
avgRev
## [1] 15.7
t$difSq <- (t$comp-avgRev)^2
var <- sum(t$difSq)
var
## [1] 370.58
sd <- sqrt(var)
sd
## [1] 19.25045
  1. About how much revenue should the airline expect for a flight of 120 passengers? With what standard deviation? Note any assumptions you make and if you think they are justified.

For a flight of 120 passengers, the airline should expect $15.7 * 120 = $1,884 in revenue. The variance is Var(aX) = a^2 x Var(X) where a = 120 and Var(X) was found in (a). The standard deviation here is thus 2310.06.

sqrt(120^2 * var)
## [1] 2310.055

2.44 Income and gender. The relative frequency table below displays the distribution of annual total personal income (in 2009 inflation-adjusted dollars) for a representative sample of 96,420,486 Americans. These data come from the American Community Survey for 2005-2009. This sample is comprised of 59% males and 41% females.

  1. Describe the distribution of total personal income. The distribution is right-skewed as few individuals with high total personal income bring the mean above the median.

  2. What is the probability that a randomly chosen US resident makes less than $50,000 per year?

.212+.183+.158+.047+.022
## [1] 0.622
  1. What is the probability that a randomly chosen US resident makes less than $50,000 per year and is female? Note any assumptions you make.

Here, I am assuming that gender and income are independent. So, P(is female and makes less than 50,000)

0.622*.41
## [1] 0.25502
  1. The same data source indicates that 71.8% of females make less than $50,000 per year. Use this value to determine whether or not the assumption you made in part (c) is valid.

Using this information, the assumption that I made in part (c) is not valid. This is saying that P(makes less than 50,000 | is female) = .718. Since P(makes less than 50,000) = .622 != .718, P(A|B) != P(A), so income and gender are not independent.