install.packages("data.tree")suppressWarnings(suppressMessages(library(knitr)))HW6 1. A box contains 54 red marbles, 9 white marbles, and 75 blue marbles. If a marble is randomly selected from the box, what is the probability that it is red or blue? Express your answer as a fraction or a decimal number rounded to four decimal places.
Solution.
The probabilithy that get a red marble or blue marble from the box is:
54/(54+9+75)+75/(54+9+75) = 0.9348Solution.
The probability that you end up with a red golf ball is:
20/(19+20+24+17) = 0.2500
Gender and Residence of Customers Males Females Apartment 81 228 Dorm 116 79 With Parent(s) 215 252 Sorority/Fraternity House 130 97 Other 129 72
v1 <- c("Apartment","Dorm","With Parent(s)","Sorority/Fraternity House","Other")
Males <- c(81,116,215,130,129)
Females <- c(228,79,252,97,72)
df <- data.frame(v1,Males,Females)
colnames(df) <- c("","Males","Females")
kable(df, format = "pandoc",full_width = F,caption = "Gender and Residence of Customers", position = "left")| Males | Females | |
|---|---|---|
| Apartment | 81 | 228 |
| Dorm | 116 | 79 |
| With Parent(s) | 215 | 252 |
| Sorority/Fraternity House | 130 | 97 |
| Other | 129 | 72 |
What is the probability that a customer is not male or does not live with parents? Write your answer as a fraction or a decimal number rounded to four decimal places.
Solution.
total attended the surway:
81+116+215+130+129+228+79+252+97+72=1399
The probability that a customer is not male or does not live with parents is:
(228+79+252+97+72)/1399 + (215+252)/1399 = 0.8542
4.Determine if the following events are independent. Going to the gym. Losing weight.
Answer: A) Dependent B) Independent
B)
Discussion.
The events A and B are independent if any one of the following three equivalent conditions hold:
P(A ??? B) = P(A)P(B)
P(A|B) = P(A) B has no effect on A
P(B|A) = P(B) A has no effect on B
Let G denote Going to gym, the sample space of whether a person going to gym or not looks like this:
{Y,N}
Let L be the event that the person is losing weight or not. Then event L looks like this:
{y,n}
The sample space of the person going to gym and then losing weight looks like this:
{Yy, Yn,Ny, Nn}
The sample space of the person losing weight and then going to gym looks like this:
{yY, yN, nY, nN}
If each event has the same probability, then
P(G|L) = {yY,nN} = 2/4 = 1/2
P(L|G) = {Yy,Nn} = 2/4 = 1/2
If the probability of the person going to gym P(G) = 1/2 ,then P(G|L) = P(G). Then we know G and L are independent. If
P(G) does not equal to 1/2, then P(G|L) does not equal to P(G) and G and L are dependent. Or, if each event has different probability so P(G|L) does not equal to 1/2, then P(G|L) does not equal to P(G) and G and L are dependent.
If the probability of the person losing weight P(L) = 1/2 ,then P(L|G) = P(L). Then we know G and L are independent. If
P(L) does not equal to 1/2, then P(L|G) does not equal to P(L) and G and L are dependent. Or, if each event has different probability so P(L|G) does not equal to 1/2, then P(L|G) does not equal to P(L) and G and L are dependent.
# Solution
c8_3 <- factorial(8)/(factorial(5)*factorial(3))
c7_3 <- factorial(7)/(factorial(4)*factorial(3))
var <- c8_3 * c7_3 * 3
cat("Answer: There are",var,"veggie wraps can be made.")## Answer: There are 5880 veggie wraps can be made.6.Determine if the following events are independent. Jeff runs out of gas on the way to work. Liz watches the evening news.
Answer: A) Dependent B) Independent
Answer: B)7.The newly elected president needs to decide the remaining 8 spots available in the cabinet he/she is appointing. If there are 14 eligible candidates for these positions (where rank matters), how many different ways can the members of the cabinet be appointed?
#Solution
# The permutation of 14 people taking 8 at a time is:
p14_8 <- factorial(14)/factorial(6)
cat("Answer: There are",p14_8,"ways that the members of the
cabinet can be appointed.")## Answer: There are 121080960 ways that the members of the
## cabinet can be appointed.8.A bag contains 9 red, 4 orange, and 9 green jellybeans. What is the probability of reaching into the bag and randomly withdrawing 4 jellybeans such that the number of red ones is 0, the number of orange ones is 1, and the number of green ones is 3? Write your answer as a fraction or a decimal number rounded to four decimal places.
#Solution
c4_1 <- factorial(4)/(factorial(3)*factorial(1))
c9_3 <- factorial(9)/(factorial(6)*factorial(3))
cat("the probability is ",c4_1*c9_3/((9+4+9)^4))## the probability is 0.0014343289.Evaluate the following expression. \(\frac{11!}{7!}\)
factorial(11)/factorial(7)## [1] 792010.Describe the complement of the given event. 67% of subscribers to a fitness magazine are over the age of 34.
The complement of the event could be expressed as: 33% of the subscribers to a fitness magazine are not older than 34.Step 1. Find the expected value of the proposition. Round your answer to two decimal places.
Step 2. If you played this game 559 times how much would you expect to win or lose? (Losses must be entered as negative.)
#Solution
#Random Number values are +97 and -30.
#Probabilities to get three head out of 4 tosses is 4C3/2^4 =1/4 and the complement is 1-1/4=3/4.
E <- round(((1/4)*97 - (3/4)*30),2)
print(paste0("I can expect to win $", E," each time I play the game."))## [1] "I can expect to win $1.75 each time I play the game."print(paste0("If I played this game 559 times, I expect to win $", E*559, "."))## [1] "If I played this game 559 times, I expect to win $978.25."Step 1. Find the expected value of the proposition. Round your answer to two decimal places.
Step 2. If you played this game 994 times how much would you expect to win or lose? (Losses must be entered as negative.)
#Solution
#Random Number values are +23 and -26.
#Probabilities to get 4 tails or less out of 9 tosses is (9C4+9C3+9C2+9C1)/2^9 and the complement is 1-9P4/2^9.
p <- (factorial(9)/(factorial(5)*factorial(4))+factorial(9)/(factorial(6)*factorial(3))+factorial(9)/(factorial(7)*factorial(2))+factorial(9)/(factorial(8)*factorial(1))+factorial(9)/(factorial(9)*factorial(0)))/(2^9)
E <- 23*p -26*(1-p)
print(paste0("I can expect to lose $", abs(E)," each time I play the game."))## [1] "I can expect to lose $1.5 each time I play the game."print(paste0("If I played this game 994 times, I expect to lose $", abs(E*994), "."))## [1] "If I played this game 994 times, I expect to lose $1491."suppressWarnings(suppressMessages(library(data.tree)))
truth <- Node$new("True")
True <- truth$AddChild("True")
Lie <- truth$AddChild("Lie")
truth$probability <- '0.8'
True$probability <- '0.8*0.9'
Lie$probability <- '0.8*(1-0.9)'
liar <- Node$new("Lie")
True <- liar$AddChild("True")
Lie <- liar$AddChild("Lie")
truth$probability <- '0.2'
True$probability <- '0.2*(1-0.59)'
Lie$probability <- '0.2*0.59'
print(truth,"probability")## levelName probability
## 1 True 0.2
## 2 ¦--True 0.8*0.9
## 3 °--Lie 0.8*(1-0.9)print(liar,"probability")## levelName probability
## 1 Lie
## 2 ¦--True 0.2*(1-0.59)
## 3 °--Lie 0.2*0.59# Probability is P (is a ture liar|detected as a liar)
# bayes theorem: P(A|B) = (P(B|A)P(A))/P(B)
#probability of detected as a liar: P(B)= 0.8*0.1+0.2*0.59
#probability of actually a liar: P(A)=0.2
#probability of given actually a liar, detected as a liar: P(B|A)= 0.2*0.59
p_ab<-(0.2*0.59)*0.2/(0.8*0.1+0.2*0.59)
cat("P(A|B) =",p_ab)## P(A|B) = 0.1191919# Probability is P (is a ture truth-teller|detected as a truth-teller)
# bayes theorem: P(A|B) = (P(B|A)P(A))/P(B)
#probability of detected as a truth-teller: P(B)= 0.8*0.9+0.2*(1-0.59)
#probability of actually a truth-teller: P(A)=0.8
#probability of given actually a truth-teller, detected as a truth-teller: P(B|A)= 0.8*0.9
p_ab<-(0.8*0.9)*0.8/(0.8*0.9+0.2*(1-0.59))
cat("P(A|B) =",p_ab)## P(A|B) = 0.7182045# Probability is P (is a ture truth-teller|detected as a truth-teller)
#probability of truth teller detected as a liar: P(B)= 0.8*(1-0.9)
#probability of liar detected as a liar: P(A)=0.2*0.59
p_aorb<- (0.8*(1-0.9))+(0.2*0.59)
cat("P(A|B) =",p_aorb)## P(A|B) = 0.198