Solution

Let’s represent as follows:

Defining function that find the probability of a color based on total quantity!

P <- function(C, Total){
  p <- C / Total
  return(p)
}

Since the marble is randomly selected from the box and we want to know the probability of red or blue, we have as follows:

\[P(R \cup B) = P(R) + P(B)\] \(P(R \cup B)\) = 0.3913 + 0.5435

\(P(R \cup B)\) = 0.9348

Answer: The probability that the marble is red or blue is 0.9348 or 93.48%.

Solution

Let’s represent as follows:

Defining function that find the probability of a color based on total quantity!

P <- function(C, Total){
  p <- C / Total
  return(p)
}

Since the marble is randomly selected from the machine and we want to know the probability of red, we have as follows:

\[P(R) = \frac{\#\:Red\:golf\:balls}{Total\:number\:of\:golf\:balls\:in\:the\:machine}\]

\(P(R)\) = 0.25

Answer: The probability that the golf ball is a red is 0.25 or 25%.

Solution

Let’s calculate our totals as follows:

Since we are being asked the probability that a customer is not male or does not live with parents, we have as follows:

\(P(Not\:Male) = 1 - P(Male)\)

\(P(Not\:living\:with\:parents)\) = 1 - \(P(living \: with \: parents)\)

\(P(Not\:Male \: \cup \: Not \: living \: With \: Parents) = (1 - P(Male)) + (1 - P(living\:with\:parents)) - (1 - P(Not \: Male \: \cap \: Not \: living\:With\:Parents\)

Answer: The probability that a customer is not male or does not live with parents is 0.5327 or 53.27%.

Solution

1. Going to the gym.

Answer: Independent

2. Losing weight.

Answer: Dependent

Solution

We can choose 56 different combinations for our vegetables.

We can choose 35 different combinations for our condiments.

We can choose 3 different combinations for our tortillas.

Answer: The wrap can be made in 5880 different ways.

Solution

Jeff runs out of gas on the way to work.

Answer: Dependent

Liz watches the evening news.

Answer: Independent

Solution

Table of positions:

Answer: The members of the cabinet can be appointed in 121080960 different ways; but since rank matters the number of ways will be 3003 different ways.

Solution

Let’s represent as follows:

Defining function that find the probability of a color based on total quantity!

Since we have to withdraw 4 jelly beans, the resulting probability of the output will be as follows:

Individual probability:

\(p = P(O) \times P(G) \times P(G) \times P(G)\)

Combined probability:

#p <- pO * pG * pG * pG # Individual probabilities
p <- ( choose(R, 0) * choose(O, 1) * choose(G, 3) ) / choose(R + O + G, 4) # Combined probabilities

Answer: 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 is 0.0459 or 4.59%.

Solution

\[\frac{11!}{7!}= \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}\]

\[\frac{11!}{7!}= 11 \times 10 \times 9 \times 8\]

Answer: \(\frac{11!}{7!}\) = 7920

Solution

\(p(Age > 34) = 0.67\)

\(p(Age \le 34) = 1 - p(Age > 34)\)

\(p(Age \le 34) = 1 - 0.67\)

\(p(Age \le 34) = 0.33\)

Answer: \(33\%\) of subscribers to a fitness magazine are under or equal the age of \(34\).

Solution

1. Find the expected value of the proposition. Round your answer to two decimal places.

By assuming that it is a fair coin and by applying Bernoulli, we have as follows:

n <- 4 # Number of trials
j <- 3 # Number of success times
p <- 1/2   # Probability of winning on every trial is the same
q <- 1 - p   # Probability of failure on every trial is the same
dollar_win <- 97 # Dollar amount if wining
dollar_loose <- -30  # Dollar amount if loosing

Bernoulli formula.

bernoulli <- function(n,p,j){
  b <- choose(n,j) * p ^ (j) * (1 - p) ^ (n - j)
  return(b)
}

b <- bernoulli(n,p,j)
# dbinom(j,n,p) # Binomial distribution function from R.

The probability of getting exactly three heads is 0.25.

Now, in order to calculate our dollar values, we can do as follows:

Winning

\(E(Win) = bernoulli(n,p,j) \times dollar\_win\)

\(E(Win) = 0.25 \times 97\)

\(E(Win)\) = 24.25

Loosing

\(E(Loose) = (1 - bernoulli(n,p,j)) \times dollar\_loose\)

\(E(Loose) = 0.75 \times -30\)

\(E(Loose)\) = -22.5

Answer: Based on the above results and the given dollar values, the expected amount to be made is higher than the amount to be paid out.

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

\(E(Pay) = E(Win) + E(Loose)\)

ntimes <- 559 
EWin <- bernoulli(n,p,j) * dollar_win * ntimes
ELoose <- (1 - bernoulli(n,p,j)) * dollar_loose * ntimes
Expected_Pay <- EWin + ELoose

Answer: After 559 times I would expect to win $13555.75; loose $-12577.5 and end up with a profit of $978.25.

Solution

1. Find the expected value of the proposition. Round your answer to two decimal places.

By assuming that it is a fair coin and by applying Bernoulli, we have as follows:

n <- 9 # Number of trials
j <- 4 # Number of success times must be equal to or less than 4
p <- 1/2   # Probability of winning on every trial is the same
q <- 1 - p   # Probability of failure on every trial is the same
dollar_win <- 23 # Dollar amount if wining
dollar_loose <- -26  # Dollar amount if loosing

Bernoulli formula.

bernoulli <- function(n,p,j){
  b <- choose(n,j) * p ^ (j) * (1 - p) ^ (n - j)
  return(b)
}

b0 <- bernoulli(n,p,0) # Probability of 0
b1 <- bernoulli(n,p,1) # Probability of 1
b2 <- bernoulli(n,p,2) # Probability of 2
b3 <- bernoulli(n,p,3) # Probability of 3
b4 <- bernoulli(n,p,4) # Probability of 4
b <- b0 + b1 + b2 + b3 + b4
#pbinom(j,n,p) # Binomial distribution function from R.

The probability of getting 4 tails or less is 0.5.

Now, in order to calculate our dollar values, we can do as follows:

Winning

\[E(Win) = dollar\_win \times \sum{bernoulli(n,p,j_i)}\]

\[E(Win) = 23 \times 0.5 \]

\(E(Win)\) = 11.5

Loosing

\[E(Loose) = dollar\_loose \times (1 - \sum{bernoulli(n,p,j_i)}) \]

\[E(Loose) = -26 \times 0.5 \]

\(E(Loose)\) = -13

Answer: Based on the above results and the given dollar values, the expected amount to be made is lower than the amount to be paid out.

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

\(E(Pay) = E(Win) + E(Loose)\)

ntimes <- 994 
EWin <- b * dollar_win * ntimes
ELoose <- (1 - b) * dollar_loose * ntimes
Expected_Pay <- EWin + ELoose

Answer: After 994 times I would expect to win $11431; loose $-12922 and end up with a loss of $-1491.

Solution

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.)

pLiarGivenLiar <- liars_table$`Tell Lies`[1] / liars_table$Total[1]

Answer: The probability that an individual is actually a liar given that the polygraph detected him/her as such is 0.596 or 59.6%.

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.)

pTruthGivenTruth <- liars_table$`Tell Truth`[2] / liars_table$Total[2]

Answer: The probability that an individual is actually a truth-teller given that the polygraph detected him/her as such is 0.8978 or 89.78%.

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.

pliar <- liars_table$`Tell Lies`[3] +  liars_table$Total[1] - liars_table$`Tell Lies`[1]

Answer: The probability that a randomly selected individual is either a liar or was identified as a liar by the polygraph is 0.28 or 28%.

END.