red <- 54
white <- 9
blue <- 75
prob_red_or_blue = round((red + blue) / (red + white + blue), 4)
prob_red_or_blue## [1] 0.9348The probability of randomly selecting a red or blue marble from the box is 0.9348.
green <- 19
red <- 20
blue <- 24
yellow <- 17
prob_red <- round(red / (green + red + blue + yellow), 4)The probability of randomly getting a red golf ball is 0.25.
Gender and Residence of Customers
| Location of residence | Males | Females |
|---|---|---|
| Apartment | 81 | 228 |
| Dorm | 116 | 79 |
| With Parent(s) | 215 | 252 |
| Sorority/Fraternity House | 130 | 97 |
| Other | 129 | 72 |
First I did this the complicated way below…
prob_not_male <- 1-((81 + 116 + 215 + 130 + 129) / 1399)
prob_not_lives_w_parents <- 1-((215 + 252) / 1399)
prob_not_both <- prob_not_male + prob_not_lives_w_parents - (1-215/1399)
prob_either_or <- prob_not_male + prob_not_lives_w_parents - prob_not_both
prob_either_or## [1] 0.8463188Then I realized that the probability that a customer is not male OR that they do not live with parents is the same as the inverse of the probability that they are male AND live with their parents…
prob_male_and_lives_w_parents <- 215 / 1399
prob_not_male_or_not_lives_w_parents <- round((1-prob_male_and_lives_w_parents), 4)
prob_not_male_or_not_lives_w_parents## [1] 0.8463The probability that a customer is not male or does not live with parents is 0.8463.
The two events are dependent because the probability of losing weight will presumably change based on whether or not you are going to the gym.
vegetables <- 8
condiments <- 7
tortillas <- 3
veggie_combos <- choose(vegetables, 3)
condiment_combos <- choose(condiments, 3)
wraps <- veggie_combos * condiment_combos * tortillas
wraps## [1] 5880There are 5880 different veggie wraps can be made.
The events that Jeff runs out of gas on the way to work and Liz watches the evening news are independent because the outcome of one in no way influences the probability of the other.
permutations <- function(n, r){
return (factorial(n)/factorial(n - r))
}
cabinet <- permutations(14, 8)
cabinet## [1] 121080960There are 121080960 different ways the president can appoint the remaining 8 spots in his cabinet.
red <- 9
orange <- 4
green <- 9
total <- red + orange + green
prob_1orange_3green <- round((choose(red, 0) * choose(orange, 1) * choose(green, 3)) / choose(total,4), 4)
prob_1orange_3green## [1] 0.0459The probability of choosing 0 red, 1 orange and 3 green jelly beans is 0.0459.
ans <- factorial(11)/factorial(7)
ans## [1] 7920ans2 <- 11*10*9*8
ans2## [1] 7920\[ \frac{11!}{7!} = 7920 \]
Since 67% of subscribers are over the age of 34 the probability of a randomly chosen subscriber being over the age of 34 is 0.67 and the complement of that probability would be 1 - 0.67 or 0.33 probability that a randomly chosen subscriber is 34 or younger.
prob_3_heads <- pbinom(3, size=4, prob=0.5) - pbinom(2, size=4, prob=0.5)
prob_3_heads## [1] 0.25comp <- 1 - prob_3_heads
ex_v <- round(prob_3_heads*97 - comp*30, 2)
ex_v## [1] 1.75The expected value of the proposition is 1.75.
winnings <- 559 * (ex_v)
winnings## [1] 978.25I would expect to win 978.25 dollars.
prob_4_or_less_tails <- pbinom(4, size=9, prob=0.5)
prob_4_or_less_tails## [1] 0.5comp <- 1 - prob_4_or_less_tails
ex_v <- round(prob_4_or_less_tails*23 - comp*26, 2)
ex_v## [1] -1.5The expected value of the proposition is -1.5.
losses <- 994 * (ex_v)
losses## [1] -1491I would expect to lose -1491 dollars.
liar <- 0.2
truth_teller <- 0.80
liar_sensitivity <- 0.59
truth_teller_sensitivity <- 0.90
p_pos_det_liar <- liar * liar_sensitivity
p_false_det_liar <- truth_teller * (1-truth_teller_sensitivity)
p_actually_liar <- p_pos_det_liar/(p_pos_det_liar + p_false_det_liar)
p_actually_liar## [1] 0.5959596There is a 0.596 probability that an individual is actually a liar when the polygraph identifies them as a liar.
p_pos_det_truth <- truth_teller * truth_teller_sensitivity
p_false_det_truth <- liar * (1-liar_sensitivity)
p_actually_truth_teller <- p_pos_det_truth/(p_pos_det_truth + p_false_det_truth)
p_actually_truth_teller## [1] 0.8977556There is a 0.8978 probability that an individual is actually a “truth teller” when the polygraph identifies them as a “truth teller”.
liars_or_identified <- liar + p_false_det_liar
liars_or_identified## [1] 0.28There is a 0.28 probability that a randomly selected individual is actually a liar whether they are detected or not, or identified as a liar even though they are telling the truth.