red = 54
white = 9
blue = 75
success01 = red+blue
total01 = red+white+blue
prob01 = success01/total01
roundprob01 = round(prob01,4)
roundprob01## [1] 0.9348\[P(Red \cup Blue) = \frac{Num(Red)+Num(Blue)}{Num(Red)+Num(White)+Num(Blue)} = \frac{54+75}{54+9+75} = \frac{43}{46} = 0.9348\]
green = 19
red = 20
blue = 24
yellow = 17
success02 = red
total02 = green+red+blue+yellow
prob02 = success02/total02
prob02## [1] 0.25\[P(R) = \frac{Num(Red)}{Num(Green)+Num(Red)+Num(Blue)+Num(Yellow)} = \frac{20}{19+20+24+17} = \frac{1}{4} = 0.25\]
matr03 %>% kable(caption="**Gender and Residence of Customers**") %>%
kable_styling(c("striped", "bordered"))| 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.
By De Morgan’s law, \({\left(\neg Male \right) \vee \left(\neg WithParents \right) = \neg \left( Male \wedge WithParents \right)}\)
So:
totl = sum(matr03)
Males=sum(matr03[,"Males"])
notMales=totl - Males
WithParents = sum(matr03["With Parent(s)",])
notWithParents = totl - WithParents
MaleAndWithParents = matr03["With Parent(s)","Males"]
notMaleOrNotWithParents = totl - MaleAndWithParents
result = round(notMaleOrNotWithParents/totl,4)## Total: 1399## Total Males: 671## Not Males: 728## With Parents: 467## Not With Parents: 932## Male AND With Parents: 215## NotMale OR NotWithParents: 1184## result: 0.8463\[{Pr[\left(\neg Male \right) \vee \left(\neg WithParents \right)] = 1- Pr[\left( Male \wedge WithParents \right)]}\\ =1-\frac{215}{1399}=\frac{1184}{1399}=0.8463\]
By Bayes Rule, \[Pr(A|B)=\frac{Pr(A \wedge B)}{Pr(B)}\] .
If the events are independent, \[Pr(A \wedge B) = Pr(A) \cdot Pr(B)\] , so for independent events, \[Pr(A|B) = \frac{Pr(A \wedge B)}{Pr(B)} = \frac{Pr(A) \cdot Pr(B)}{Pr(B)}=Pr(A)\] To be independent, \[Pr(Going to Gym \wedge Losing Weight) = Pr(Going to Gym)\cdot Pr(Losing Weight)\]
Alternatively, the events are independent if \[Pr(Losing Weight | Going to Gym) = Pr(Losing Weight)\]
However, a person is more likely to lose weight if he/she does go to the gyn, i.e. \[Pr(Losing Weight | Going to Gym) > Pr(Losing Weight)\]
Therefore, these events are NOT independent.
Answer: A) Dependent B) Independent
vegetablechoice = choose(8,3)
condimentchoice = choose(7,3)
tortillachoice = choose(3,1)
result = vegetablechoice * condimentchoice * tortillachoice
result## [1] 5880There are 56 ways to choose 3 vegetables from 8 and there are 35 ways to choose 3 condiments from 7.
Therefore the number of possible veggie wraps is \(56 \cdot 35 \cdot 3 = 5880\) .
There does not appear to be any relationship between these two events, especially as they are involving separate people.
Answer: A) Dependent B) Independent
cabinet be appointed?
Because order matters, we need to use permutations, \(P(n,k) = \frac{n!}{(n-k)!}\), rather than combinations, \(C(n,k) = \frac{n!}{k!(n-k)!}\).
permute <- function(n,k) {
factorial(n) / factorial(n-k)
}
candidates = 14
slots = 8
result = permute(candidates,slots)
#result
fmtresult = format(result,big.mark=",", trim=TRUE)
fmtresult## [1] "121,080,960"The number of permutations to fill 8 from 14, where sequence matters, is \[\frac{14!}{(14 - 8)!} = \frac{14!}{(6)!} = 121,080,960\]
totalRED = 9
totalORANGE = 4
totalGREEN = 9
totalJELLYBEANS = totalRED + totalORANGE + totalGREEN
chooseRED = 0
chooseORANGE = 1
chooseGREEN = 3
chooseJELLYBEANS = chooseRED + chooseORANGE + chooseGREEN
combRED = choose(totalRED,chooseRED)
combORANGE = choose(totalORANGE,chooseORANGE)
combGREEN = choose(totalGREEN,chooseGREEN)
combSELECTED = combRED * combORANGE * combGREEN
combJELLYBEANS = choose(totalJELLYBEANS,chooseJELLYBEANS)
result = round(combSELECTED / combJELLYBEANS,4)
result## [1] 0.0459There is 1 way to choose 0 Red jellybeans from 9 (i.e., choose none of them.)
There are 4 ways to choose 1 Orange jellybeans from 4.
There are 84 ways to choose 3 Green jellybeans from 9.
Thus, the number of ways to select the above combination is \(1 \cdot 4 \cdot 84 = 336\) .
There are 7315 ways to choose 4 jellybeans from 22 total jellybeans.
Therefore the probability of selecting the desired combination is \(\frac{336}{7315} = 0.0459\).
## [1] 7920The answer is \(\frac{11!}{7!}=\frac {11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} {7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = {11 \cdot 10 \cdot 9 \cdot 8} = 7920\) .
The event is \(Age(Subscriber)>34\) ,
and the probability of the event is \(Pr(Age(Subscriber)>34)=0.67\) .
In other words: choose a random subscriber to the fitness magazine.
The probability that the randomly selected individual is older than 34 is 0.67 .
The complement of the event is \(Age(Subscriber) \le 34\),
and the probability of the complement is \(Pr(Age(Subscriber) \le 34) = 1 - Pr(Age(Subscriber)>34)=0.67 = 0.33\) .
In other words, the probability that a randomly selected subscriber to the magazine is age 34 or younger is 0.33 .
There are four ways to get exactly three heads in 4 tosses: {HHHT,HHTH,HTHH,THHH}.
The number of possible results in 4 tosses is \(2^4=16\). Therefore, the probability that I win 97 dollars is \(Pr(Win) = \frac{4}{16}=\frac{1}{4}=0.25\) , and the probability that I lose 30 dollars is \(Pr(Lose) = 1-Pr(Win) = 1-0.25 = 0.75\) .
winget = 97
losspay = -30
waystowin = choose(4,3)
eventspace = 2^4
waystolose = eventspace - waystowin
probwin = waystowin / eventspace
#probwin
problose = waystolose / eventspace
#problose
expected = round(probwin * winget + problose * losspay,2)
#expectedThe expected value of the game is \[Pr(Win)\cdot Payoff(Win) + Pr(Lose) \cdot Payoff(Lose) = 0.25 \cdot \$97 - 0.75 \cdot \$30 = 1.75\]
The expected result of playing 559 times is \(\$1.75 \cdot 559 = \$978.25\).
successtails = (0:4)
successes = array(choose(9,successtails),dimnames=list(paste(successtails,"tail(s)")))
#successes
numsuccesses = sum(successes)
#numsuccesses
totalpossibilities = 2^9
#totalpossibilities
probsuccess = numsuccesses / totalpossibilities
#probsuccess
probfailure = 1 - probsuccess
#probfailure
winget = 23
losepay = -26
expectedgame = winget * probsuccess + losepay*probfailure
#expectedgame| 0 tail(s) | 1 tail(s) | 2 tail(s) | 3 tail(s) | 4 tail(s) |
|---|---|---|---|---|
| 1 | 9 | 36 | 84 | 126 |
and the sum of this table is 256. The total number of possible results from 9 tosses is \(2^9 = 512\),
so the probability of winning is \(\frac{256}{512} = 0.5\) and the probability of losing is \(1 - 0.5 = 0.5\) .
The expected result from one play of this game is thus \[0.5 \cdot 23 \textrm{ dollars} + 0.5 \cdot (-26)\textrm{ dollars} = -1.5\textrm{ dollars} \] .
If this game were played 994 times, the expected result would be $-1491 dollars, i.e., you would expect to lose $1491 .
In the analysis, “Positive” indicates a liar, while “Negative” indicates a Truthteller.
grid = matrix(c("TP","FP","TP+FP",
"FN","TN","FN+TN",
"TP+FN","FP+TN","TP+FN+FP+FN"),
3,3,byrow=T,
dimnames = list(c("PredictedLiar","PredictedTruthTeller","TotalPredicted"),
c("ActualLiar","ActualTruthTeller","TotalActual")))
grid %>% kable() %>% kable_styling(c("striped", "bordered"))| ActualLiar | ActualTruthTeller | TotalActual | |
|---|---|---|---|
| PredictedLiar | TP | FP | TP+FP |
| PredictedTruthTeller | FN | TN | FN+TN |
| TotalPredicted | TP+FN | FP+TN | TP+FN+FP+FN |
Since 20 percent of the individuals are actually Liars, \(TP+FN=0.2\) , which means \(FP+TN=1-0.2=0.8\) .
grid["TotalPredicted","ActualLiar"] = paste(grid["TotalPredicted",
"ActualLiar"],"=0.200")
grid["TotalPredicted","ActualTruthTeller"] = paste(grid["TotalPredicted",
"ActualTruthTeller"],"=0.800")
grid["TotalPredicted","TotalActual"] = paste(grid["TotalPredicted",
"TotalActual"],"=1.000")
grid %>% kable() %>% kable_styling(c("striped", "bordered"))| ActualLiar | ActualTruthTeller | TotalActual | |
|---|---|---|---|
| PredictedLiar | TP | FP | TP+FP |
| PredictedTruthTeller | FN | TN | FN+TN |
| TotalPredicted | TP+FN =0.200 | FP+TN =0.800 | TP+FN+FP+FN =1.000 |
\[ \begin{aligned} \textrm{Sensitivity} &= \frac{\textrm{TruePositives}}{\textrm{TruePositives+FalseNegatives}} \\ &=\frac{TP}{TP+FN} = \frac{TP}{0.200} = 0.59 \\ &=\frac{\textrm{ActualLiars}}{\textrm{ActualLiars+MislabeledTruthtellers}} \end{aligned} \]
Therefore, \(TP = (TP+FN) \cdot 0.59 = 0.200 \cdot 0.59 = 0.118\) ,
and \(FN = (TP+FN)-TP = 0.200 - 0.118 = 0.082\) .
grid["PredictedLiar","ActualLiar"] = paste(grid["PredictedLiar",
"ActualLiar"],"=0.118")
TP=0.118
grid["PredictedTruthTeller","ActualLiar"] = paste(grid["PredictedTruthTeller",
"ActualLiar"],"=0.082")
FN=0.082
grid %>% kable() %>% kable_styling(c("striped", "bordered"))| ActualLiar | ActualTruthTeller | TotalActual | |
|---|---|---|---|
| PredictedLiar | TP =0.118 | FP | TP+FP |
| PredictedTruthTeller | FN =0.082 | TN | FN+TN |
| TotalPredicted | TP+FN =0.200 | FP+TN =0.800 | TP+FN+FP+FN =1.000 |
\[ \begin{aligned} \textrm{Specificity} &= \frac{\textrm{TrueNegatives}}{\textrm{FalsePositives+TrueNegatives}} \\ &=\frac{TN}{FP+TN} = \frac{TN}{0.800} = 0.90 \\ &=\frac{\textrm{ActualTruthTellers}}{\textrm{MislabeledLiars+ActualTruthtellers}} \end{aligned} \]
Therefore, \(TN = (FP+TN) \cdot 0.90 = 0.800 \cdot 0.90 = 0.720\) ,
and \(FP = (FP+TN)-TN = 0.800 - 0.720 = 0.080\) .
grid["PredictedLiar","ActualTruthTeller"] = paste(grid["PredictedLiar",
"ActualTruthTeller"],"=0.080")
FP=0.080
grid["PredictedTruthTeller","ActualTruthTeller"] = paste(grid["PredictedTruthTeller",
"ActualTruthTeller"],"=0.720")
TN=0.720
grid %>% kable() %>% kable_styling(c("striped", "bordered"))| ActualLiar | ActualTruthTeller | TotalActual | |
|---|---|---|---|
| PredictedLiar | TP =0.118 | FP =0.080 | TP+FP |
| PredictedTruthTeller | FN =0.082 | TN =0.720 | FN+TN |
| TotalPredicted | TP+FN =0.200 | FP+TN =0.800 | TP+FN+FP+FN =1.000 |
Completing the rightmost totals in the grid:
grid["PredictedLiar","TotalActual"] = paste(grid["PredictedLiar",
"TotalActual"],"=0.198")
grid["PredictedTruthTeller","TotalActual"] = paste(grid["PredictedTruthTeller",
"TotalActual"],"=0.802")
grid %>% kable() %>% kable_styling(c("striped", "bordered"))| ActualLiar | ActualTruthTeller | TotalActual | |
|---|---|---|---|
| PredictedLiar | TP =0.118 | FP =0.080 | TP+FP =0.198 |
| PredictedTruthTeller | FN =0.082 | TN =0.720 | FN+TN =0.802 |
| TotalPredicted | TP+FN =0.200 | FP+TN =0.800 | TP+FN+FP+FN =1.000 |
Confirm that these results give the correct metrics:
## [1] 0.59## [1] TRUE## [1] 0.9## [1] TRUEThe percentage which are actually liars is TP =0.118 .
The percentage predicted to be liars is TP+FP =0.198 .
## [1] 0.596The probability that an individual is actually a liar, given that the polygraph detected him/her as such is 0.596 .
This metric is known as precision or positive predictive value.
The percentage which are actually truthtellers is TN =0.720 .
The percentage predicted to be truthtellers is FN+TN =0.802 .
## [1] 0.8978The probability that an individual is actually a truth-teller, given that the polygraph detected him/her as such, is 0.8978.
This metric is known as negative predictive value.
By DeMorgan’s Law,
\[ \begin{aligned} Pr(ActualLiar \vee PredictedLiar)&= 1 - Pr(\neg(ActualLiar) \wedge \neg(PredictedLiar)\\ &= 1-Pr(ActualTruthTeller \wedge PredictedTruthTeller) \\ &= 1 - 0.72 \\ &= 0.28 \end{aligned} \]