Week 2 Bridge Math HW
Monu Chacko
January 2, 2019
Problem 1. Dice Rolls
If you roll a pair of fair dice, what is the probability of..
(a) getting a sum of 1?
(b) getting a sum of 5?
(c) getting a sum of 12?
Answer Problem 1. Dice Rolls
There are 36 total possibilities
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
(a) getting a sum of 1?
The probability of getting a sum of 1 is 0 out of 36
P(1) = \(\frac{0}{36} = 0\)%
(b) getting a sum of 5?
The probability of getting a sum of 5 is 4 out of 36 or 11.11%
P(5) = \(\frac{4}{36} = 11.11\)%
(c) getting a sum of 12?
The probability of getting a sum of 12 is 1 out of 36 or 2.78%
P(12) = \(\frac{1}{36} = 2.78\)%
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.
Answer Problem 2. School absences
(a) What is the probability that a student chosen at random doesn’t miss any days of school due to sickness this year?
P(1|S) = 25% or .25
P(2|S) = 15% or .15
P(>=3|S) = 28% or .28
P(S) = 0.25 + 0.15 + 0.28 = 0.68
Probability of a student not missing due to sickness = 1 - (0.25 + 0.15 + 0.28) = 0.32 or 32%
(b) What is the probability that a student chosen at random misses no more than one day?
P(2|S) = 15% or .15
P(>=3|S) = 28% or .28
Probability of a student missing no more than one day = 1 - (0.15 + 0.28) = 0.57 or 57%
(c) What is the probability that a student chosen at random misses at least one day?
Probability that a student misses at least one day = 0.25 + 0.15 + 0.28 = 0.68 or 68%
(d) 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.
Probability of Child 1 not missing school = 0.32
Probability of Child 2 not missing school = 0.32
Total Probability = (0.32 x 0.32) = 0.1024 or 10.24%
(e) 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.
Probability of Child 1 missing school = 0.68
Probability of Child 2 missing school = 0.68
Total Probability = (0.68 x 0.68) = 0.4624 or 46.24%
(f) 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.
If part (d) and part (e) are independent events then the numbers would be correct. If they are not (since they are siblings) then the numbers might not be accurate.
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 0.0289Answer Problem 3. Health coverage, relative frequencies
(a) Are being in excellent health and having health coverage mutually exclusive?
No, they are mutually exclusive events. You can have excellent health and have coverage at the same time.
(b) What is the probability that a randomly chosen individual has excellent health?
\[ P(E) = \frac{ 0.023 + 0.2099}{1} = \frac{0.2329}{1} = 0.2329 \]
pe = (0.023 + 0.2099) / 1
print(pe)## [1] 0.2329(c) What is the probability that a randomly chosen individual has excellent health given that he has health coverage?
\[ P(E|C) = \frac{P(C \cap E)}{P(C)} \]
E = Excellent Health
C = Has Health Coverage
\[ P(E|C) = \frac{0.2099}{0.8738} = 0.240215152 \] or 24.02%
(d) What is the probability that a randomly chosen individual has excellent health given that he doesn’t have health coverage?
\[ P(E|NC) = \frac{(P(NC \cap E)}{P(NC)} \]
E = Excellent Health
NC = Has No Health Coverage
\[ P(E|NC) = \frac{0.023}{0.1263} = 0.182106097\] or 18.21%
(e) Do having excellent health and having health coverage appear to be independent?
Excellent health and having health coverage are not dependent. However from the graph below we can see that there seems to be some coorelation between a person having health insurance and not having one. This is evident from the green line getting closer to the red line towards the end.
P(E) should be equal to P(E|I) \[ P(E) = \frac{0.2099} {0.2329} = 0.90124517 \] (The denominator 0.2329 is the sum of all excellent health i.e. 0.023 + 0.2099)
\[ P(I) = \frac{0.2329} {0.2329} = 1 \]
\[ P(E|I) = \frac{P(E \cap I)} {P(E)} \]
\[ P(E|I) = \frac{0.90124517 * 1} {0.90124517} = 1 \]
colNoHealthInsurance = c(0.023,0.0364,0.0427,0.0192,0.005)
colHasHealthInsurance = c(0.2099,0.3123,0.241,0.0817,0.0289)
dfHealth <- data.frame(colNoHealthInsurance,colHasHealthInsurance)
dfHealth <- as.data.frame(t(as.matrix(dfHealth)))
names(dfHealth) <- c("E","VG","G","F","P")
plot(colHasHealthInsurance, type="l", col = "green", xlab="Green: Covered; Red: Not Covered", ylab="Health")
lines(colNoHealthInsurance, type="l", col = "red")
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?
Answer Problem 4. Exit Poll.
Voted for Scott Walker
\[ P(SW) = 0.53 \]
Voted for other than Scott Walker
\[ P(SW) = 1 - 0.53 = 0.47 \]
Voted Scott Walker given college degree
\[ P(SW|C) = \frac{P(SW \cap C)}{1} \] \[ P(SW|C) = \frac{0.53 * 0.37}{1} = 0.1961 \]
Voted for other than Scott Walker given college degree
\[ P(Other|C) = \frac{P(Other \cap C)}{1} \] \[ P(Other|C) = \frac{0.47 * 0.44}{1} = 0.2068 \]
Has college degree and voted for Scott Walker
\[ P(C|SW) = \frac{0.1961}{(0.1961 + 0.2068)} = 0.486721271 \]
There is about 48.67% chance that a randomly sampled person with a college degree voted for Scott Walker.
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 8Answer Problem 5. Books on a bookshelf
Hard Paper Total
fiction 13 59 72
nonfiction 15 8 23
Total 28 67 95
(a) Find the probability of drawing a hardcover book first then a paperback fiction book second when drawing without replacement.
\[ P(H) = \frac{28}{95} = 0.294736842 \]
H - Hard Cover; P - Paper; F - Fiction; NF - Non Fiction
\[ P(PF|H) = \frac{59}{95-1} = \frac{59}{94} = 0.627659574 \]
\[ \frac{0.294736842}{0.627659574} = 0.184994401 \]
or 18.50%
(b) Determine the probability of drawing a fiction book first and then a hardcover book second,when drawing without replacement.
\[ P(F) = \frac{72}{95} * \frac{28}{94} = 0.225755879 \]
or 22.58%
(c) 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.
\[ P(F) = \frac{72}{95} * \frac{28}{95} = 0.223379501 \]
or 22.34%
(d) The final answers to parts (b) and (c) are very similar. Explain why this is the case.
Since there are large number of books (95 books), replacing a book and not replacing one makes little impact. If the total number of books were less, then it would have some impact.
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.
Answer Problem 6. Is it worth it?
(a) Create a probability model and find Andy’s expected profit per game.
\[ P(Num) = \frac{36} {52} = 0.692307692 \] \[ P(JKQ) = \frac{12} {52} = 0.230769231 \] \[ P(A) = \frac{4} {52} = 0.076923077 \]
\[ P(A|C) = \frac{1} {52} = 0.019230769 \]
\[ (0.692307692 * 0) + (0.230769231 * 3) + (0.076923077 * 5) + (0.019230769 * 20) = 1.461538458 \]
Andy is likely to make $1.46 per play
(b) Would you recommend this game to Andy as a good way to make money? Explain.
No, I will not recommend this play because Andy is spending $2 and getting about $1.46 back thus incurring a loss of about 0.54 per play.
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 0.0625Answer Problem 7. Scooping ice cream.
(a) 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?
1 box 48 ounces + 3 scoop from box 2 would be 6 ounces (3 x 2) = Total: 54 fluid ounces
Variance Formula:
\[ S^2 = \frac{ \sum_{} (x_{i} - \overline{x})^2}{n - 1} \]
Standard Deviation
\[ S = \sqrt{S^2} \]
Since we know the SD and Var we can calculate
var_1 = (1 + (0.25*3))
print(var_1)## [1] 1.75sd_1 = sqrt(var_1)
print(sd_1)## [1] 1.322876(b) 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?
If you take 1 scoop from a box then there would be 46 ounces (48 - 2) remaining. The Expected Value E(X) or arithmetic mean would still be 2. The SD of remaining would be
sd_r = sqrt(1-.0625)
print(sd_r)## [1] 0.9682458