Problem 1. Dice Rolls If you roll a pair of fair dice, what is the probability of..

  1. getting a sum of 1?

    probability is 0

  2. getting a sum of 5?

    probability is 0.11111111

  3. getting a sum of 12?

    probability is 0.02777778

—requires dice package—

library(dice)
## Loading required package: gtools
getSumProbs(ndicePerRoll = 2,
nsidesPerDie = 6)
## $probabilities
##       Sum Probability Ways to Roll
##  [1,]   2  0.02777778            1
##  [2,]   3  0.05555556            2
##  [3,]   4  0.08333333            3
##  [4,]   5  0.11111111            4
##  [5,]   6  0.13888889            5
##  [6,]   7  0.16666667            6
##  [7,]   8  0.13888889            5
##  [8,]   9  0.11111111            4
##  [9,]  10  0.08333333            3
## [10,]  11  0.05555556            2
## [11,]  12  0.02777778            1
## 
## $average
## [1] 7

Problem 2. School absences Data collected at elementary schools in DeKalb County, GA suggest that each year roughly 25% of students miss exactly one day of school, 15% miss 2 days, and 28% miss 3 or more days due to sickness.

  1. What is the probability that a student chosen at random doesn’t miss any days of school due to sickness this year?
x = 1 -(.25+.15+.28)
x
## [1] 0.32
  1. What is the probability that a student chosen at random misses no more than one day?
x = .32+.25
x
## [1] 0.57
  1. What is the probability that a student chosen at random misses at least one day?
x = .25+.15+.28
x
## [1] 0.68

  1. If a parent has two kids at a DeKalb County elementary school, what is the probability that neither kid will miss any school? Note any assumption you must make to answer this question.

    Assuming independent

x = .32*.32
x
## [1] 0.1024

  1. If a parent has two kids at a DeKalb County elementary school, what is the probability that both kids will miss some school, i.e. at least one day? Note any assumption you make.

    Assuming independent

x = .68*.68
x
## [1] 0.4624

  1. If you made an assumption in part (d) or (e), do you think it was reasonable? If you didn’t make any assumptions, double check your earlier answers.

    Might not be reasonable to make the assumption in (d) and (e) since sickness is the cause of missing school, sickness can spread and be contagious. Most likely the two children live in the same household.

Problem 3. Health coverage, relative frequencies The Behavioral Risk Factor Surveillance System (BRFSS) is an annual telephone survey designed to identify risk factors in the adult population and report emerging health trends. The following table displays the distribution of health status of respondents to this survey (excellent, very good, good, fair, poor) and whether or not they have health insurance.

mat=matrix(c(.023, 0.0364, 0.0427, 0.0192, 0.0050,0.2099, 0.3123 ,0.2410 ,0.0817,0.0289), byrow=TRUE, nrow=2)
colnames(mat)=c("Excellent", "Very Good","Good", "Fair","Poor")
rownames(mat)=c("No Coverage","Coverage")
mat
##             Excellent Very Good   Good   Fair   Poor
## No Coverage    0.0230    0.0364 0.0427 0.0192 0.0050
## Coverage       0.2099    0.3123 0.2410 0.0817 0.0289

  1. Are being in excellent health and having health coverage mutually exclusive?

    No

  2. What is the probability that a randomly chosen individual has excellent health?

x = .0230+.2099
x
## [1] 0.2329
  1. What is the probability that a randomly chosen individual has excellent health given that he has health coverage?
x = .2099/sum(mat[2,])
x
## [1] 0.2402152
  1. What is the probability that a randomly chosen individual has excellent health given that he doesn’t have health coverage?
x = .0230/sum(mat[1,])
x
## [1] 0.1821061

  1. Do having excellent health and having health coverage appear to be independent?

    No

Problem 4. Exit Poll. Edison Research gathered exit poll results from several sources for the Wisconsin recall election of Scott Walker. They found that 53% of the respondents voted in favor of Scott Walker. Additionally, they estimated that of those who did vote in favor for Scott Walker, 37% had a college degree, while 44% of those who voted against Scott Walker had a college degree. Suppose we randomly sampled a person who participated in the exit poll and found that he had a college degree. What is the probability that he voted in favor of Scott Walker?

mat=matrix(c(.1961, 0.3339, 0.2068, 0.2632), byrow=TRUE, nrow=2)
colnames(mat)=c("Degree", "NoDegree")
rownames(mat)=c("VoteWalker","NoVoteWalker")
mat
##              Degree NoDegree
## VoteWalker   0.1961   0.3339
## NoVoteWalker 0.2068   0.2632
x =.1961/sum(mat[,1])
x
## [1] 0.4867213

Problem 5. 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.

mymat2=matrix(c(13,59,15,8),nrow=2,byrow=TRUE)
colnames(mymat2)=c("hard","paper")
rownames(mymat2)=c("fiction","nonfiction")
mymat2
##            hard paper
## fiction      13    59
## nonfiction   15     8
  1. Find the probability of drawing a hardcover book first then a paperback fiction book second when drawing without replacement.
x = sum(mymat2[,1]/sum(mymat2))*(59/(sum(mymat2)-1))
x
## [1] 0.1849944
  1. Determine the probability of drawing a fiction book first and then a hardcover book second,when drawing without replacement.
x = sum(mymat2[1,]/sum(mymat2))*(sum(mymat2[,1]/(sum(mymat2)-1)))
x
## [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.
x = sum(mymat2[1,]/sum(mymat2))*sum(mymat2[,1])/sum(mymat2)
x
## [1] 0.2233795

  1. The final answers to parts (b) and (c) are very similar. Explain why this is the case.

    Total books difference was only 1 book.

Problem 6. Is it worth it? Andy is always looking for ways to make money fast. Lately, he has been trying to make money by gambling. Here is the game he is considering playing: The game costs 2 dollars to play. He draws a card from a deck. If he gets a number card (2-10), he wins nothing. For any face card (jack, queen or king), he wins 3 dollars. For any ace, he wins 5 dollars and he wins an extra $20 if he draws the ace of clubs.

  1. Create a probability model and find Andy’s expected profit per game.
x = (36/52*-2)+(12/52*1)+(3/52*3)+(1/52*23)
x
## [1] -0.5384615

  1. Would you recommend this game to Andy as a good way to make money? Explain.

    No, there is money to be made from this.

Problem 7. Scooping ice cream. Ice cream usually comes in 1.5 quart boxes (48 fluid ounces), and ice cream scoops hold about 2 ounces. However, there is some variability in the amount of ice cream in a box as well as the amount of ice cream scooped out. We represent the amount of ice cream in the box as X and the amount scooped out as Y . Suppose these random variables have the following means, standard deviations, and variances:

mymat3=matrix(c(48,1,1, 2,.25,.0625), nrow=2, byrow=TRUE)
colnames(mymat3)=c("mean", "SD", "Var")
rownames(mymat3)=c("X, In Box","Y, Scooped")
mymat3
##            mean   SD    Var
## X, In Box    48 1.00 1.0000
## Y, Scooped    2 0.25 0.0625
  1. An entire box of ice cream, plus 3 scoops from a second box is served at a party. How much ice cream do you expect to have been served at this party? What is the standard deviation of the amount of ice cream served?
x = 48+3*2
x
## [1] 54
x = sqrt(1+.0625+.0625+.0625)
x
## [1] 1.089725
  1. How much ice cream would you expect to be left in the box after scooping out one scoop of ice cream? That is, find the expected value of X ??? Y . What is the standard deviation of the amount left in the box?
x = 48-2
x
## [1] 46
x = sqrt(1+.0625)
x
## [1] 1.030776

  1. Using the context of this exercise, explain why we add variances when we subtract one random variable from another.

    overall variability is still increased