HW6 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.

Answer:

red_marble <- 54
white_marble <- 9
blue_marble <- 75

round((red_marble+blue_marble)/(red_marble+blue_marble+white_marble),4)
## [1] 0.9348

The probability of it’s red or blue is 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.

Answer:

green_ball <- 19
red_ball <- 20
blue_ball <- 24
yellow_ball <- 17

total <- green_ball + red_ball + blue_ball + yellow_ball
round( red_ball/total,4)
## [1] 0.25

The probability o fbeing red is 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
data.frame(Housing= c("Apartment","Dorm","With Parent(s)","Sorority/Fraternity House","Other"),Males= c(81,116,215,130,129),Females= c(228,79,252,97,72))
##                     Housing Males Females
## 1                 Apartment    81     228
## 2                      Dorm   116      79
## 3            With Parent(s)   215     252
## 4 Sorority/Fraternity House   130      97
## 5                     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.

Answer:

not_male <- 228 + 79 + 252 + 97 + 72
not_live_parent <- 81 + 228 + 116 + 79 + 130 + 97 + 129 + 72
not_male_parent <- 228 + 79 + 97 + 72
total <- 81 + 116 + 215 + 130 + 129 + 228 + 79 + 252 + 97 + 72
ans <- not_male/total + not_live_parent/total - not_male_parent/total
round(ans, 4)
## [1] 0.8463

P(not male or not live with parent) = P(not male)+P(not female) - P(not male and not female) = 0.8463 The probability of not male or female is 0.8463.

  1. Determine if the following events are independent. Going to the gym. Losing weight. Answer: A) Dependent

  2. 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?

Answer: The answer will be multplication of the combination of vegetables, condiments and tortilla.

choose(8,3)*choose(7,3)*choose(3,1)
## [1] 5880

6.Determine if the following events are independent. Jeff runs out of gas on the way to work. Liz watches the evening news.

Answer: B) 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?

Answer: 14 permutation 8 will be the answer.

factorial(14)/factorial(14-8)
## [1] 121080960

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.

Answer:

favouravle <- choose(9,0)*choose(4,1)*choose(9,3)
total_outcome <- choose(22,4)
round(favouravle/total_outcome,4)
## [1] 0.0459
  1. Evaluate the following expression. \(\frac{11!}{7!}\) Answer:
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.

Answer: 33% of subscribers to a fitness magzaine are below the age of 34 years.

11.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.

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

Answer: The coin toss follows binomial distribution with size 4, probability 0.5 and the number of wins 3. Step 1

win_prob <- dbinom(3, size = 4, prob = 0.5)
loss_prob <- 1 -win_prob
win_prob * 97 + loss_prob*-30
## [1] 1.75

Expected value to win is 1.75.

Step 2 Playing 559 times the expected win is

559 * 1.75
## [1] 978.25
  1. Flip a coin 9 times. If you get 4 tails or less, I will pay you $23. Otherwise you pay me $26. 12. 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.)

Answer: Step 1.

win_prob <- dbinom(4, size = 9, prob = 0.5) +  dbinom(3, size = 9, prob = 0.5)+ dbinom(2, size = 9, prob = 0.5)+ dbinom(1, size = 9, prob = 0.5)+ dbinom(0, size = 9, prob = 0.5)
loss_prob <- 1 -win_prob
expected <- win_prob * 23 + loss_prob*-26
expected
## [1] -1.5

The expected value is -1.5. Step 2

expected * 994
## [1] -1491

The expected loss of 994 time playing is -1491.

  1. 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.

  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.)

  2. 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.)

  3. 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.

Answer:

sensitivity <- 0.59
specificiity <- 0.90
liar <- 0.2
(sensitivity * liar)/(sensitivity*liar + (1-specificiity)*(1-liar))
## [1] 0.5959596
(specificiity * (1-liar))/((1-sensitivity)*liar+ specificiity*(1-liar))
## [1] 0.8977556
  1. P(liar or identified as liar) = P(liar) + P(identified as liar) - P(liar and identified as liar) = 0.2 + 0.59 - 0.59*0.12
0.2 + 0.59 - 0.59*0.2
## [1] 0.672

The probability is 0.672.