The probabiltiy that the randomly selected marble is either red or blue is (54+75)/(54+75+9)=0.9348.
The probability that you end up with a red golf ball is 20/(19+20+24+17)=.25
customers <- matrix(c(81, 228, 116, 79, 215, 252, 130, 97, 129, 72),ncol=2,byrow=TRUE)
colnames(customers) <- c("Males","Females")
rownames(customers) <- c("Apartment","Dorm","With Parents", "Sorority/Fraternity House", "Other")
customers <- as.table(customers)
customers
## Males Females
## Apartment 81 228
## Dorm 116 79
## With Parents 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.
The probability that a customer is not male or does not live with parents is (1399-215)/1399=0.8463
The following events are independent. There is probably a correlation between the two, but one doesn’t necessarily cause the other.
x <- (choose(3, 1))*(choose(7, 3))*(choose(8, 3))
x
## [1] 5880
There are 5880 different veggie wraps.
The events are independent.
y <- 14*13*12*11*10*9*8*7
y
## [1] 121080960
There are 121,080,960 different ways the members of the cabinet can be appointed.
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.
Evaluate the following expression. (11!)/(7!)=(11)(10)(9)(8)=7920
10, Describe the complement of the given event. 67% of subscribers to a fitness magazine are over the age of 34.
33% of subscribers to a fitness magazine are 34 years old and younger.
(97)(.25)+(-30)(.75)=1.75
The expected value is $1.75
Step 2. If you played this game 559 times how much would you expect to win or lose? (Losses must be entered as negative.)
559*1.75= 978.25. I’d expect to win $978.25
(23)(.5)+(-26)(.5)= -1.55
The expected value is $1.75
Step 2. If you played this game 994 times how much would you expect to win or lose? (Losses must be entered as negative.)
994*(-1.5)= -1491. I’d expect to lose $1491.
Define P(L) to be the probability that someone is lying. Define P(T) to be the probability that someone is not lying. Define P(PL) to be the probability that the test detects lying. Define P(PN) to be the probability that the test doesn’t detect lying.
P(L|PL)=( (P(L))(P(PL|L)) )/P(PL) = ( (P(L))(P(PL|L)) )/(((P(L))P(PL|L)) + ((P(T))P(PL|T))) = (.2)(.59)/((.2)(.59)+ (.8)(.1))=.596
P(T|PN)=((P(T))(P(PN|T)))/P(PN) = ((P(T))(P(PN|T)))/(((P(T))(P(PN|T))) + ((P(L))P(PN|L))) = (.8)(.9)/((.9)(.9)+ (.2)(.41))=.898
P(L ∪ PL)= P(L)+P(PL)-P(L ∩ PL)= .2 + ((.2)(.59)+ (.8)(.1)) - (.2)(.59) = .28