Week 2 Homework

2.6 Dice Rolls

  1. Since there are 6 sides on a die, and there is a pair of dice being rolled, the possible outcomes are 6 x 6 = 36. However, because the lowest number on a die is 1, the lowest possible sum when rolling a pair of dice is 2.Therefore, the probability of getting a sum of 1 when rolling a pair of die is 0/36 or just 0.
  2. There are 4 possible outcomes that result in a sum of 5: (1,4) (2,3) (3,2) (4,1). Therefore the probability of getting a sum of 5 when rolling a pair of dice is 4/36 or 11.1%.
  3. There is just one outcome that results in a sum of 12: (6,6). The probability of getting a sum of 12 is 1/36 or 2.8%.

2.12 School Absences

  1. The probability that a student did not miss any days of school due to sickness can be found subtracting the probability a student will miss AT LEAST one day of school due to sickness from 1. The probability that a student will miss AT LEAST one day of school due to sickness is found by adding the probabilities for the student missing 1 day (25%), 2 days (15%), and 3 or more days (28%), equaling 68%. Therefore, the probability that a student chosen at random didn’t miss any days of school due to sickness is 32%.
  2. The probability that a student misses no more than one day is found by adding the probability that a student misses 0 days (32%) and 1 day (25%) to equal 57%.
  3. The probability that a student misses at least one day was found previously, by adding the probabilities of missing 1,2 and 3 or more days due to sickness. The probability was 68% that a student will miss at least one day.
  4. To find the probability that two kids of the same parent will both not miss any school, we must use the multiplication rule for independent events. The probability of a student not missing any school is 32%. Since the events are independent of each other, we multiply the probabilities together to get .32 x .32 = .1024 or 10.24%. (For this problem I assumed the events to be independent)
  5. The probability that two kids of the same parent will both miss at least one day of school is found by multiplying the probabilities of each kid missing at least one day of school to get .68 x .68 = .4624 or 46.24%. (For this problem I assumed the events to be independent)
  6. I think it was reasonable to make the assumption that the events in d) and e) were both independent. However, you could make the argument that if the parent is a single or stay-at-home parent, then the probability one of their kids going to school could be affected if the other kid is not able to go to school because the parent may not be able to take the other kid to school and therefore keeping them both home.

2.18 Health Coverage, Relative Frequencies

  1. No. Having excellent health and having health insurance are not mutually exclusive because they can occur at the same time.
  2. The probability that a randomly chosen individual has excellent health is found by adding the probabilities for people with excellent health with and without health insurance. The probability is .2329 or 23.29%.
  3. The probability an individual has excellent health given that he has health coverage is .2099 or 20.99%.
  4. The probability an individual has excellent health given that he does not have health coverage is .0230 or 2.3%
  5. Having excellent health and having health insurance are not independent because having health insurance changes the probability of having excellent health.

2.24 Exit Poll

To solve this problem we must first find some probabilities. The probability that a person has a college degree AND voted for Scott Walker is .37 x .53 = .1961 or 19.61%. The probability a person has a college degree AND voted against Scott Walker is .44 x .47 = .2068 or 20.68%. Now, we use the Law of Conditional probability to find the probability that a person voted for Scott Walker, given that the have a college degree. Let B = the event that voter has a college degree Let A1 = the event that they voted for Scott Walker P(B|A1) = probability of having college degree, given voted for Scott Walker = .37 P(A1) = probability voted for Scott Walker = .53 P (B|A2) = probability of having a college degree, given voted against Scott Walker = .44 P(A2)= probability voted against Scott Walker = .47

P(A1|B) = (.37)(.53) / [(.37)(.53) + (.44)(.47)] = .4867 = 48.67%

The probability that a person with a college degree voted for Scott Walker is 48.67%.

2.30 Books on a Bookshelf

  1. The probability of picking a hardcover book first and then a paperback fiction second without replacement is found by multiplying the probability of picking a hardcover book times the probability of picking a paperback fiction, after the hardcover is already removed from the total. This gives us (28/95) x (59/94) = .1849 or 18.49%.
  2. To find the probability of picking a fiction book first and a hardcover second, we multiply the probabilities without replacement. This gives us (72/95) x (28/94) = .2258 or 22.58%
  3. This time the probability would be (72/95) x (28/95) = .2233 or 22.33%
  4. The final answers to b) and c) are very similar in this case because there are a large number of books on the shelf so removing/replacing one book does not change the probability much

2.36 Is it worth it?

  1. P(A1) = 36/52 → $-2 P(A2) = 12/52 → $1 P(A3) = 3/52 → $3 P(A4) = 1/52 → $23 We can find the Andy’s expected profit per game by adding the expected profits for each card multiplied by its probability of being picked…

P = (-2)(36/52) + 1(12/52) + 3(3/52) + 23(1/52)

= -1.3077 + .2308 + .1731 + .4423 = -.4615

  1. No. This game is not a good way of making money because on average you will lose $.46 each game you play.

2.42 Scooping Ice Cream

  1. I would expect 54 ounces of ice cream to be served because one full box is 48 ounces and 3 scoops hold about 2 ounces. The standard deviation of the ice cream served is equal to the square root or the sum of the squared variances of a full box and 3 scoops. Therefore, sqrt(1 + .25 +.25 +. 25) = 1.3229.
  2. The expected value of X – Y would just be 48 – 2 = 48. The standard deviation of the amount left in the box would be sqrt(1 + .25) = 1.1180
  3. We must add the variances even though we subtracting one variable from another because any change, whether it is more or less, to the box of ice cream creates more uncertainty about how much ice cream is left. Therefore, even if we are taking out ice cream, we must add the variances because there is actually more room for error.