Solution 1:
P(R∪B) = N(R)+N(B) / N(T) = (54+75)/(54+75+9) = 0.9348
Solution 2:
P(R) = N(R)/N(T) = 20/(19+20+24+17) = 0.25
Gender and Residence of Customers
Residence 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 3:
1 - P(male and live with parents)
1 - (215)/(81+116+215+130+129+228+79+252+97+72)
## [1] 0.8463188
Going to the gym. Losing weight. Answer: Dependent
Solution 4:
An independent event is an event that has no connection to another event’s chances of happening (or not happening). In other words, the event has no effect on the probability of another event occurring. Hence, Going to the gym. Losing weight. are dependent events as going to the gym increase the probability of losing weight.
Solution 5:
C(n,k) = n!/(n-k)!k!
(factorial(8)/(factorial(3)*factorial(5)))*(factorial(7)/(factorial(3)*factorial(4)))*(factorial(3)/(factorial(2)*factorial(1)))
## [1] 5880
Jeff runs out of gas on the way to work. Liz watches the evening news.
Solution 6:
Independent An independent event is an event that has no connection to another event’s chances of happening (or not happening). In other words, the event has no effect on the probability of another event occurring. Hence, Jeff runs out of gas on the way to work. Liz watches the evening news. are independent evens as “Jeff runs out of gas on the way to work” has no effect on the probability of another event occurring.
Solution 7:
P(n,k)=n!(n−k)!
factorial(14)/factorial(14-8)
## [1] 121080960
Solution 8:
choose(9,0)*choose(4,1)*choose(9,3)/choose(22,4)
## [1] 0.04593301
Solution 9:
11*10*9*8
## [1] 7920
67% of subscribers to a fitness magazine are over the age of 34.
Solution 10:
33% of subscribers to a fitness magazine are 34 or younger.
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 11:
p.win<-choose(4,3)*((1/2)^3)*((1/2)^1)
p.win
## [1] 0.25
P(win) = 0.25
P(loss) = 1 - P(win) = 0.75
Step 1: 970.25 + 0.75(-30) = 1.75
Step 2: 559 * 1.75 = 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 12:
p.win<-pbinom(4,9,1/2)
p.win
## [1] 0.5
P(win) = 0.5
P(loss) = 1 - P(win) = 0.5
Step 1: 23p.win-26(1-p.win) = -1.5
Step 2: 994* (-1.5) = -1491
a <- c('Result', 'Liar', 'Truth-Teller','Total')
b <- c('+', 0.59*0.2, 0.1*0.8, 0.59*0.2+0.1*0.8)
c <- c('-', 0.41*0.2, 0.9*0.8, 0.41*0.2+0.9*0.8)
d <- c('Total', 0.59*0.2+0.41*0.2, 0.1*0.8+0.9*0.8, (0.59*0.2+0.1*0.8+0.41*0.2+0.9*0.8))
t(data.frame(a,b,c,d))
## [,1] [,2] [,3] [,4]
## a "Result" "Liar" "Truth-Teller" "Total"
## b "+" "0.118" "0.08" "0.198"
## c "-" "0.082" "0.72" "0.802"
## d "Total" "0.2" "0.8" "1"
0.59*0.2/(0.59*0.2+0.1*0.8)
## [1] 0.5959596
0.9*0.8/(0.41*0.2+0.9*0.8)
## [1] 0.8977556
P(Liar)+P(Liar_polygraph/“Truth−Teller”)
0.2+0.1*0.8
## [1] 0.28