Prob1a : Rolling 2 fair Dices and getting sum of 1

answer: You will always be getting greater than 1

# require(dice)
# 
# my.sum <- readline(prompt="The Sum You like to use: ")
# my.sum <- as.integer(my.sum)
# print(paste("You entered,", my.sum))
# 
# two.dice <- function(){
#   dice <- sample(1:6, size = 2, replace = TRUE)
#   return(sum(dice))
# }
# 
# s <-two.dice()
# s
# if(s==my.sum)
# {
#    print(sprintf("The sum is %s", my.sum))
#    
#    prob = (my.sum-1)/36
#   
#    print(sprintf("The prob is %s", prob))
#  }else
#  {
#    print(sprintf("The sum is %s", s))
#    
#    prob = (my.sum-1)/36
#    
#    print(sprintf("The prob is  %s", prob))
#  }
#

Prob 1b,c: Rolling 2 fair Dices and getting sum of 5 & sum of 12 using 100,000 repetition for “long run average”

1b, answer: For sum of 5, based on 100000 repittions, the prob of getting a sum of 5 is 0.11

1c, answer:

For sum of 12, based on 100000 repittions, the prob of getting a sum of 5 is 0.0269

## Loading required package: dice
## Loading required package: gtools
## sims
##       2       3       4       5       6       7       8       9      10 
## 0.02753 0.05598 0.08414 0.11123 0.13803 0.16525 0.13986 0.11047 0.08431 
##      11      12 
## 0.05464 0.02856

Probability prob statment: 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.

prob 2.1:

P(1)= 0.25 P(2)=0.15 P(>=3)=0.28

  1. What is the probability that a student chosen at random doesn’t miss any days of school due to sickness this year?

P(0)+P(1)+P(2)+P(>=3)=1.0 Therefore, P(0) = 1-0.25-0.15-0.28 = 0.32

Prob2.2: (b) What is the probability that a student chosen at random misses no more than one day?

P(1)+P(2) =0.32+0.25=0.57

Prob2.3: (c) What is the probability that a student chosen at random misses at least one day?

P(1)+P(2)+P(>=3)=0.25+0.15+0.28=0.68

Prob2.4: (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.

Let A be one of the kids and B be the other kid

P(A)and P(B)= P(0)XP(0)=0.32*0.32 = 0.1024

This assumes the two children getting sick (events) are independent

Prob2,5: (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.

From part C, P(at least one day missing)=0.68

P(A)and P(B)= P(at least 1 day missing)*P(at least 1 day missing)=0.4624

This assumes the two children getting sick (events) are independent

  1. The assumption is reasonable because one child getting sick has not dependencies to the other child

Prob 3:

  1. Are being in excellent health and having health coverage mutually exclusive? No,

P(excellent)=0.023+0.2099=0.2329 P(health coverage)=0.2099+0.3123+0.2410+0.0817+0.0289=0.8738 For Not mutually exclusive:

PP(excellent and health coverage) <> zero

P(excellent or health coverage)=P(excellent)+P(health coverage)-P(excellent and health coverage) = 0.8738

  1. What is the probability that a randomly chosen individual has excellent health? 0.023+0.2099 =0.2329

  2. What is the probability that a randomly chosen individual has excellent health given that he has health coverage? P(excellent and health coverage)=0.2099 P(coverage) = 0.8738

P(excellent|health coverage)=P(excellent and health coverage)/P(coverage) = 0.2099/0.8738= 0.2665

  1. What is the probability that a randomly chosen individual has excellent health given that he doesn’t have health coverage?

P(excellent and no coverage) = 0.0230 P(no coverage) = 0.1263

P(excellent|No health coverage)=P(excellent and no coverage)/P(no coverage)

= 0.0230/0.1263=0.1821

  1. Do having excellent health and having health coverage appear to be independent? No, they are dependent events

From part c: P(excellent|health coverage)=P(excellent and health coverage)/P(coverage) = 0.2099/0.8738= 0.266

P(health coverag|excellent)=P(excellent and health coverage)/P(excellent)

=0.2009/0.2329=0.8626

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

Prob 4: Exit poll conditional probablity

P(voted scott)= 0.53 P(against scott)=1-0.53=0.47 (degree|Scott’) = 0.44

P(scott|degree)=P(degree|scott)P(Scott)/(P(degree|scott)P(Scott)+(P(degree|scott’)*P(Scott’))

=(0.37)(0.53)/(0.37)(0.53)+(0.44)*(0.47)

=0.48672

Prob 5: (a) Find the probability of drawing a hardcover book first then a paperback fiction book second when drawing without replacement. (28/95) * (59/94) = .185

  1. Determine the probability of drawing a fiction book first and then a hardcover book second,when drawing without replacement.

(72/95) * (28/94) = .2258

  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.

(72/95) * (28/95) = .2234

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

The probability with and without replacement is very small because the no. of fiction and nonfiction books very close. So the difference in chances of drawing a fiction hardcover vs. nonfiction harcover is relatively small

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

Prob 6: (a) Create a probability model and find Andy’s expected profit per game. For $2 dollar bet

P( 2 - 10) = 36 / 52 = 9 / 13 => $0 (No profit) P(J, Q, K) = 12 / 52 = 3 / 13 => $3 ($1 profit) P(Ace) = 4 / 52 = 1 / 13 => $5 ($3 Profit) P(Ace of Clubs) = 1 / 52 => $20 ($18 Profit)

P = (-2)(9/13) + 1(3/13) + 3(1/13) + 18(1/52) -1.3846 + 0.2308 + 0.2308 + 0.3462 Expected Earnings = -0.578

  1. Would you recommend this game to Andy as a good way to make money? Explain. No, he’s expected to lose on avg. 58 cent

prob7: (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?

48+3*2=54 ounces

For independent variables: Var1+var2 +var3 +..

Therefore, Var of 1 box + var of 3 scopes = 1 + 0.06253 The Std Dev = sqrt (1+ 30.0625) =1.089

  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?

"1 Box - 1 scoop = 48 - 2=46

same reasoning as previous part, adding variances together: sqrt(1 + .0625)=1.038

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

Since each scoop is randomly sampled, whether we add or subtract from the pile, the variance increases; more variability has been introduced into the system.

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