Problem 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.

numR <- 54
numW <- 9
numB <- 75

allMarbles <- numR+numW+numB
probRorB <- (numR+numB)/allMarbles
round(probRorB, digits = 4)
## [1] 0.9348

Problem 2

You are going to play mini golf. A ball machine that contains 19 green golf balls, 20 red golf balls, 24 blue golf balls, and 17 yellow golf balls, randomly gives you your ball. What is the probability that you end up with a red golf ball? Express your answer as a simplified fraction or a decimal rounded to four decimal places.

numG <- 19
numR <- 20
numB <- 24
numY <- 17

allBalls <- numG+numR+numB+numY
probR <- numR/allBalls
round(probR, digits = 4)
## [1] 0.25

Problem 3

A pizza delivery company classifies its customers by gender and location of residence. The research department has gathered data from a random sample of 1399 customers. The data is summarized in the table below.

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.

mApt <- 81
mDorm <- 116
mPar <- 215
mFrat <- 130
mOther <- 129
allMale <- mApt + mDorm + mPar + mFrat + mOther

fApt <- 228
fDorm <- 79
fPar <- 252
fFrat <- 97
fOther <- 72
allFemale <- fApt + fDorm + fPar + fFrat + fOther

probForP <- (allFemale + allMale - mPar)/ (allMale + allFemale)

round(probForP, digits = 4)
## [1] 0.8463

Problem 4

Determine if the following events are independent: Going to the gym. Losing weight.
These events are dependent because in general, going to the gym causes an individual to lose weight.

Problem 5

A veggie wrap at City Subs is composed of 3 different vegetables and 3 different condiments wrapped up in a tortilla. If there are 8 vegetables, 7 condiments, and 3 types of tortilla available, how many different veggie wraps can be made?

The key to this problem is that our veggie and condiments cannot be repeated in the same sub. This means:

  1. Our first veggie slot will have 8 options, our second veggie slot will have 7 options (because we already chose one), and our third veggie slot will have 6 options.
  2. Similarly, our first condiment slot will have 7 options, our second will have 6, and our third will have 5.
  3. Since we only can only have 1 tortilla, we have 3 total options.

Additionally, we know that the order of the veggies/condiments does not matter, so we need to take this into account by removing this from the total number of combinations that we will calculate.

Our final number of combinations is:

numV <- 8
numC <- 7
numT <- 3
numSlotsV <- 3
numSlotsC <- 3

veggieCombos <- numV * (numV-1) * (numV - 2)
condCombos <- numC * (numC-1) * (numC-2)

totalCombos <- veggieCombos * condCombos * numT

# remove double-counting as a result of order for veggies and condiments
finalCombos <- totalCombos / (factorial(numSlotsV) * factorial(numSlotsC))

finalCombos
## [1] 5880

Problem 6

Determine if the following events are independent. Jeff runs out of gas on the way to work. Liz watches the evening news.
These events are independent because Jeff running out of gas does not affect Liz watching the evening news.

Problem 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?

Since rank matters, we know that our combination calculation should take order into account. We have 8 spots with 14 candidates, so the first slot will have 14 choices, and every slot thereafter will have 1 less choice.

nSpots <- 8
nCand <- 14

candCombos <- nCand

for (i in 1: (nSpots - 1)){
  candCombos <- candCombos * (nCand-i)
}

candCombos
## [1] 121080960
## This problem can also be completed in the following way: 
factorial(nCand)/ factorial(nCand - nSpots)
## [1] 121080960

Problem 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.

To find this probability, we need to calculate the following:
\[P(0\: red,\: 1 \: orange, \: 3 \: green) = \frac{num\: ways\: event\: can\: happen}{total\: number\: of\: jellybean\: choices}\]

The number of ways the event can happen can be broken down into 3 parts:

  1. The number of ways 0 reds can be chosen: \[{{9}\choose{0}} \]
  2. The number of ways 1 orange can be chosen: \[{{4}\choose{1}} \]
  3. The number of ways 3 greens can be chosen. \[{{9}\choose{3}} \]
nRed <- 9
nOrange <- 4
nGreen <- 9
totalBeans <- nRed + nOrange + nGreen
nChoices <- 4

totalChoices <- choose(22,4)
numWaysEvent <- choose(nRed,0)*choose(nOrange,1) * choose(nGreen,3)

round(numWaysEvent / totalChoices, digits = 4)
## [1] 0.0459

Problem 9

Evaluate the following expression: \[\frac{11!}{7!}\]

factorial(11)/factorial(7)
## [1] 7920

Problem 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 of age or younger.

Problem 11

If you throw exactly three heads in four tosses of a coin you win $97. If not, you pay me $30.

1. Find the expected value of the proposition. Round your answer to two decimal places.
We will need to find the probability of 3 heads

winAmt <- 97
loseAmt <- 30
pHeads <- 0.5

winProb <- pbinom(3,4, pHeads) - pbinom(2,4, 1- pHeads)
expWinnings <- (winProb * winAmt) - ((1-winProb)*loseAmt)

paste0('The expected winnings are: ',round(expWinnings,4))
## [1] "The expected winnings are: 1.75"

2. If you played this game 559 times how much would you expect to win or lose? (Losses must be entered as negative.)

paste0('After 559 games, we would expect to win: ', 559*round(expWinnings,4))
## [1] "After 559 games, we would expect to win: 978.25"

Problem 12

Flip a coin 9 times. If you get 4 tails or less, I will pay you $23. Otherwise you pay me $26.

1. Find the expected value of the proposition. Round your answer to two decimal places.
The expected value is the total number of tails we can expect to get in 9 tosses. We can further identify the probability of getting 4 or less tails by using the cumulative probability function. Finally, we will calculate the expected winnings:

maxTails <- 4
pTails <- 0.5
numTosses <- 9
winAmt <- 23
loseAmt <- 26

expVal <- pTails *numTosses
probWin <- pbinom(maxTails, size=numTosses, prob = pTails)
expWinnings <- (probWin * winAmt) - ((1-probWin) * loseAmt)

paste0('The expected winnings are: ',round(expWinnings,4))
## [1] "The expected winnings are: -1.5"

2. If you played this game 994 times how much would you expect to win or lose? (Losses must be entered as negative.)

paste0('After 994 games, we would expect to win: ', 994 * round(expWinnings,4))
## [1] "After 994 games, we would expect to win: -1491"

Problem 13

The sensitivity and specificity of the polygraph has been a subject of study and debate for years. A 2001 study of the use of polygraph for screening purposes suggested that the probability of detecting a liar was .59 (sensitivity) and that the probability of detecting a “truth teller” was .90 (specificity). We estimate that about 20% of individuals selected for the screening polygraph will lie.

a. What is the probability that an individual is actually a liar given that the polygraph detected him/her as such? (Show me the table or the formulaic solution or both.)
Let’s assume 100 individals take the lie detector test. We estimate that 20% of the individuals will lie (20 people), so we know that 80 will tell the truth. Of the 20 estimated people that lie, we know that 59% of them will be detected. Similarly, of those that are truth telling, we know that 90% will be detected. This will form the basis of our table.

nParticipants <- 100
pctLie <- 0.2
liars <- nParticipants * pctLie
truth <- nParticipants - liars


pctTrueLie <- 0.59
pctTrueTruth <- 0.9 

tp <- liars * pctTrueLie
fp <- liars - tp
tn <- truth * pctTrueTruth
fn<- truth - tn

finalTable <- data.frame(c(tp,fp, liars), c(tn, fn, truth), c(tp + tn,fp + fn, nParticipants))
names(finalTable) = c('Positive (Liars)', 'Negative (Truth)', 'All')
row.names(finalTable) = c('True', 'False', 'Total')

finalTable
##       Positive (Liars) Negative (Truth)   All
## True              11.8               72  83.8
## False              8.2                8  16.2
## Total             20.0               80 100.0

The final probability of a true liar is the total number of correctly classified liars over the total number of actual liars:

pctTrueLiar <- tp / (tp+fn)
pctTrueLiar
## [1] 0.5959596

b. What is the probability that an individual is actually a truth-teller given that the polygraph detected him/her as such? (Show me the table or the formulaic solution or both.)
The final probability of a true truth-teller is the total number of correctly classified truth-tellers over the total number of true truth-tellers:

pctTrueTruth <- tn / (tn+fp)
pctTrueTruth
## [1] 0.8977556

c. What is the probability that a randomly selected individual is either a liar or was identified as a liar by the polygraph? Be sure to write the probability statement.

We know that: \[P(A∪B)=P(A)+P(B)−P(A∩B)\] In other words: \[P(liar∪identified\:liar)=P(liar)+P(identified\: liar)−P(liar∩identified\:liar)\] Therefore, we know that:

(liars + (tp+fp) - tp) / nParticipants
## [1] 0.282