Data 605 Assignment #6

  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.
red = 54
white = 9
blue = 75

total = red + white + blue

prob_red_or_blue <- (red + blue)/total
round(prob_red_or_blue, 4)
## [1] 0.9348
  1. 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.
g_golf_balls = 19
r_golf_balls = 20
b_golf_balls = 24
y_golf_balls = 17

total <- g_golf_balls + r_golf_balls + b_golf_balls + y_golf_balls

red_ball_prob <- r_golf_balls/total
round(red_ball_prob, 4)
## [1] 0.25
  1. 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
pizza_frame <- data.frame(c(81,116, 215, 130, 129), c(228,79,252,97,72), c(309,195,467,227,201))
names(pizza_frame) <- c("Males", "Females", "Total")
row.names(pizza_frame) <- c("Apartment", "Dorm", "With Parents(s)", "Sorority/Fraterniy House", "Other")

pizza_frame
##                          Males Females Total
## Apartment                   81     228   309
## Dorm                       116      79   195
## With Parents(s)            215     252   467
## Sorority/Fraterniy House   130      97   227
## Other                      129      72   201

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

P_female = sum(pizza_frame$Females)/sum(pizza_frame$Total)

P_Not_Parents = 1 - (467/sum(pizza_frame$Total))

P_Female_Not_Parents = (sum(pizza_frame$Females) - 252)/sum(pizza_frame$Total)

Answer <- (P_female + P_Not_Parents) - P_Female_Not_Parents

round(Answer, 4)
## [1] 0.8463
  1. Determine if the following events are independent. Going to the gym. Losing weight.

Answer: Dependent. Going to the gym and presumably exercising is going to affect whether you lose weight.

  1. 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?
veggie_ans <- choose(8,3) * choose(7,3) * choose(3,1)
veggie_ans
## [1] 5880
  1. Determine if the following events are independent.

Jeff runs out of gas on the way to work. Liz watches the evening news.

Answer: Independent.

  1. 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?
factorial(14)/factorial(14-8)
## [1] 121080960
  1. 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.
bag_prob <- choose(9,0) * choose(4,1) * choose(9,3)
answer_jelly <- bag_prob/choose(22, 4)
round(answer_jelly, 4)
## [1] 0.0459
factorial(11)/factorial(7)
## [1] 7920

10.Describe the complement of the given event.

67% of subscribers to a fitness magazine are over the age of 34.

Complement = 1-0.67

Complement
## [1] 0.33
print("33% of the other subscribers are under or at 34 years of age.")
## [1] "33% of the other subscribers are under or at 34 years of age."
  1. If you throw exactly three heads in four tosses of a coin you win $97. If not, you pay me $30.

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

proposition_win <- dbinom(3, 4, 0.5, log = FALSE)
prop_loss <- 1 - proposition_win

expected_value <- (97*proposition_win)-(30 * prop_loss)
round(expected_value, 2)
## [1] 1.75

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

step2_answer <- expected_value * 559

step2_answer
## [1] 978.25

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

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

series <- seq(1, 4, by = 1)
prob <- dbinom(series, 9, 0.5, log = FALSE)
tot_val_prob <- round(sum(prob), 2)

losing_prob <- 1 - tot_val_prob
answer_12<- (23 * tot_val_prob) - (26 * losing_prob)
answer_12
## [1] -1.5

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

expect_win_loss <- answer_12 * 994
expect_win_loss
## [1] -1491
  1. The sensitivity and specificity of the polygraph has been a subject of study and debate for years. A 2001 study of theuse 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.

Assumption: 1,000 individuals are used

prob_liar <- .2
polygraph_liar <- .59
polygraph_truth <- .9

total_liars <- 1000 * prob_liar
total_truth_sayers <- 1000 - total_liars

polygraph_correct_liar <- total_liars * polygraph_liar
polygraph_correct_truth <- total_truth_sayers * polygraph_truth

polygraph_inc_liar <- total_liars - polygraph_correct_liar
polygraph_inc_truth <- total_truth_sayers - polygraph_correct_truth
t1 <- 118 + 720
t2 <- 82 + 80
t3 <- 200 + 800

df <- data.frame(c(polygraph_correct_liar,polygraph_inc_liar,total_liars), c(polygraph_correct_truth, polygraph_inc_truth, total_truth_sayers), c(t1, t2, t3))


names(df) <- c("Liars", "Truth", "Total")
row.names(df) <- c("Polygraph Correct", "Polygraph Incorrect", "Total")

df
##                     Liars Truth Total
## Polygraph Correct     118   720   838
## Polygraph Incorrect    82    80   162
## Total                 200   800  1000
  1. 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.)
actual_liar_prob <- df[1,1]/(df[2,2] + df[1,1]) 
actual_liar_prob
## [1] 0.5959596
  1. 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.)
actual_truth_prob <- df[1,2]/(df[1,2] + df[2,1])
actual_truth_prob
## [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?

Answer: 0.28

#Add up the real liars (detected or not) and the amount of people incorrectly identified as a liar
(polygraph_correct_liar + polygraph_inc_truth + polygraph_inc_liar)/1000
## [1] 0.28