Chapter 3 - Probability

Dice rolls. (3.6, p. 92) If you roll a pair of fair dice, what is the probability of

getting a sum of 1?
getting a sum of 5?
getting a sum of 12?

The denominator for all of the below events is 36 as rolling a dice is an independent event and rolling a pair of dice equates to the product of two independent events P(A) * P(B). Each dice has 6 outcomes, hence 6 * 6 =36.

ANSWER: (A) 0 - There is no way you can get a sum of 1 rolling two dice as the minimum sum you can obtain is 2, the outcomes of 1 and 1 on each dice.

ANSWER (B) 4/36 - This is the case of 2 + 3, obtaining 2 on one dice and 3 on another, there are two ways to obtain this - 2 on dice 1 and 3 on dice 2 and 3 on dice 1 and 2 on dice 2. The other case is obtaining 4 on dice 1 and the digit 1 on dice 2 totaling 5 and the opposite case.

ANSWER (C) 1/36 - There is only way to obtain 12 which is the outcome of 6 and 6 on dice 1 and on dice 2.

Poverty and language. (3.8, p. 93) 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.

Are living below the poverty line and speaking a foreign language at home disjoint?
Draw a Venn diagram summarizing the variables and their associated probabilities.
What percent of Americans live below the poverty line and only speak English at home?
What percent of Americans live below the poverty line or speak a foreign language at home?
What percent of Americans live above the poverty line and only speak English at home?
Is the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home?

##ANSWERS (A) Living below the poverty line and speaking a foreign language are not disjoint as we have 4.2% of the survey population falls in both categories as stated in the problem

(B) See below code and Venn diagram

(C) 10.4% - The non-overlapping area of set B

(D) Set A + Set B + overlap because of the “OR” condition = 16.5% + 4.2% + 10.4% = 31.1%

(E) This would be the complement of answer in D = 100% - 31.1% = 68.9%

(F) The two events are not independent as 16.5% * 10.4% != 4.2%

library("VennDiagram")

## Loading required package: grid

## Loading required package: futile.logger

venn.plot <- draw.pairwise.venn(14.6, 20.7, 4.2, c("Americans living below the poverty line", "Speak a foreign language at home"));
grid.draw(venn.plot);

grid.newpage();

Assortative mating. (3.18, p. 111) 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.

What is the probability that a randomly chosen male respondent or his partner has blue eyes?
What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes?
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?
Does it appear that the eye colors of male respondents and their partners are independent? Explain your reasoning.

##ANSWER (A) The probability formula is favorable outcomes/Total outcomes which will apply to all questions below. ## P(Male with blue eyes) + P(Female with blue eyes) - P(Male and female with blue eyes) = ## 114 + 108 - 78 = 144/204 ~= 0.7059 ## We subtract the male and female with blue eyes to avoid double-counting.

##ANSWER (B) This is a conditional probability conditioned on the male partner having blue eyes. ## P(A and B) = P(Male with blue eyes and Female with blue eyes) = 78/204 ## P(B) = P(Male with blue eyes) = 114/204 ## P(A and B) / P(B) = 78/114 ~= 0.6842

##ANSWER (C) This is a conditional probability conditioned on the male partner having brown eyes. ## P(A and B) = P(Male with brown eyes and Female with blue eyes) = 19/204 ## P(B) = P(Male with brown eyes) = 54/204 ## P(A and B) / P(B) = 19/54 ~= 0.3518

P(A and B) = P(Male with green eyes and Female with blue eyes) = 11/204

P(B) = P(Male with green eyes) = 36/204

P(A and B) / P(B) = 11/36 ~= 0.3055

##ANSWER (D) The eye colors of male respondents and their partners are not independent.

Books on a bookshelf. (3.26, p. 114) The table below shows the distribution of books on a bookcase based on whether they are nonfiction or fiction and hardcover or paperback.

Find the probability of drawing a hardcover book first then a paperback fiction book second when drawing without replacement.
Determine the probability of drawing a fiction book first and then a hardcover book second, when drawing without replacement.
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.
The final answers to parts (b) and (c) are very similar. Explain why this is the case.

##ANSWERS (A) P(A) = P(Picking out hardcoverbook first) = P(28/95) ## P(B) = P(Picking out paperback fiction book next) = P(59/94) ## Both are independent events hence P(A) * P(B) ~= 0.185

##ANSWERS (B) This can be thought of as two scenarios, the first scenario results in a fiction hardcover book being picked out in the first draw followed by a hardcover book next. The second scenario includes picking out a fiction paperback book followed by a second draw that picks a hardcover book.

P(A) = (13/95) * (27/94)

P(B) = (59/95) * (28/94)

P(A) + P(B) ~= 0.224

##ANSWERS (C) Scenraio B with replacement

P(A) = (72/95)

P(B) = (28/95)

P(A) * P(B) ~= 0.223

##ANSWERS (D) Large sample size nullifies the effect of removing one book. If the sample size were smaller, the effect would have been felt.

Baggage fees. (3.34, p. 124) 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.

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

’’’ ANSWERS (A) E(X) = 0.540 + (0.34 $25) + ((0.12 * $25) + (0.12 * $35)) = $15.7 Standard deviation = sqrt( (0 - 15.7)^2 * 0.54) + (25 - 15.7)^2 * 0.34) + (60 - 15.7)^2 * 0.12) ~= 19.9

E(X) = (0.541200) + (0.34120$25) + ((0.12120$25) + (0.12120$35)) = $1,884 Standard deviation = SQRT(120 * VAR(X)) = SQRT(120 * (19.9^2)) = 218.5

’’’

Income and gender. (3.38, p. 128) 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.

Describe the distribution of total personal income.
What is the probability that a randomly chosen US resident makes less than $50,000 per year?
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.
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.