Problem 1. Dice Rolls

  1. Probability of getting sum as 1 is 0 since minimum sum on single dice is 1, so min count of 2 dice would be 2. p(1) = 0

  2. getting a sum of 5? p(5) = p(1,4) + p(4,1) + p(2,3) + p(3,2) = 1/36 + 1/36 + 1/36 + 1/36 = 1/9 = .11111

  3. getting a sum of 12? p(12) = p(6,6) = 1/36 = .027777

Problem 2. School absences

p(one) = one day sick = .25 p(two) = two day sick = .15 p(3more) = three or more day sick = .28 p(noSick)

  1. What is the probability that a student chosen at random doesn’t miss any days of school due to sickness this year? p(noSick) = 1- (p(one) + p(two) + p(3more)) = 1 - (.25 + .15 + .28) = 1 - (.68) = .32

  2. What is the probability that a student chosen at random misses no more than one day? p(max1Day) = p(noSick) + p(one) = .32 + .25 = .57

  3. What is the probability that a student chosen at random misses at least one day? p(atLeastOne) = 1 - (p(noSick)) = .68

  4. 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. p(2KidsNoSick) = p(noSickKid1) * p(noSickKid2) = .32 * .32 = .1024

  5. 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. p(2KidsAtLeastOne) = p(atLeastOne) * p(atLeastOne) = .68 * .68 = .4624

Problem 3. Health coverage, relative frequencies

mat1=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(mat1)=c("Excellent", "Very Good","Good", "Fair","Poor")
rownames(mat1)=c("No Coverage","Coverage")
mat1
##             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, the probability of excellent health and coverage seem dependent, as it is different for coverage and no coverage

  2. What is the probability that a randomly chosen individual has excellent health? p(eH) = .023+.2099 = .2329

  3. What is the probability that a randomly chosen individual has excellent health given that he has health coverage? p(e|h) = p(e ^ h) / p(h) = .2099 / .8738 = .2402

  4. What is the probability that a randomly chosen individual has excellent health given that he doesn’t have health coverage? p(e | c~) = p(e ^ c~) / p(c~) = .0230/.1261 = .1821

  5. Do having excellent health and having health coverage appear to be independent? No, both numbers vary based on coverage.

Problem 4. Exit Poll

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?

p(s) = probability of scott walker support = .53 p(sc) = probaility of scott support and college = .53 * .37 = .1961 p(s~c) = probability for scott support and colleage = . 47 * .44 = .2068

p(s|c) = p(s ^ c) / p(c) = .1961 / .4029 = .48762

Problem 5. Books on a bookshelf

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.

p(h) * p(pf) = 28/95 * 59/94 = .18499

  1. Determine the probability of drawing a fiction book first and then a hardcover book second,when drawing without replacement. p(f).p(h) = 72/95 * 28/94 = .2257

  2. 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. p(f).p(h) = 72/95 * 28/95 = .2233

Problem 6. Is it worth it?

  1. Create a probability model and find Andy’s expected profit per game. p(0) = 9 cards = 94/(134) = 9 /13 p(3) = 3 cards = 3/ 13 p(5) = one ace = 1/13 p(20) = One Ace card (spade) = 1/52

Combining all probabilities, winning amount and cost of each game(2) = p(0)(0 - 2) + p(3)(3-2) + p(5)(5- 2) + p(20)(5 + 20 - 2) = 9/13 * -2 + 3/13 * 1 + 1/13 * 3 + 1/52 * 23 = -1.38 + .2307 + .2307 + .4423 = -.4763

  1. Would you recommend this game to Andy as a good way to make money? Explain. The average probability of losing is higher. So this game is not suggested.