Q1) What is the probability of rolling a sum of 12 on three rolls of six-sided dice? Express your answer as a decimal number only. Show your R code.
# Number of total possible results
results <- expand.grid(dice1 = seq(1,6), dice2 = seq(1,6),dice3 = seq(1,6))
nrow(results)## [1] 216# Total number of result that give a sum of 12
df <- list(rowSums(results))
table(df)## df
## 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
## 1 3 6 10 15 21 25 27 27 25 21 15 10 6 3 1# Probability of get a sum of 12 by throwing 3 times
TTR = 216 #Total number of results
DR = 25 # Desired number of results
print(DR/TTR) ## [1] 0.1157407Answer: The probability of getting a sum of 12 by throwing a 6 sided dice 3 times is about 11.57%.
Q2) What is the probability that a customer is male and lives in ‘Other’ or is female and lives in ‘Other’?
# Matrix replicaton
input <- c(200,300,200,100,100,200,200,100,200,100)
matrix <- matrix(input,nrow=5,ncol=2,byrow=TRUE)
rownames(matrix) = c("Apartment","Dorm","With Parent(s)", "Sorority/Fraternity House", "Other")
colnames(matrix) = c("males", "females")
matrix## males females
## Apartment 200 300
## Dorm 200 100
## With Parent(s) 100 200
## Sorority/Fraternity House 200 100
## Other 200 100m_other <- matrix[5,1]
f_other <- matrix[5,2]
total <- sum(matrix)
prob <- (m_other/total) + (f_other/total)
print(prob)## [1] 0.1764706Check:
# Variable values
m_other <- 200
f_other <- 100
total <- 1700
# Caculation
prob <- (m_other / total) + (f_other / total)
print(prob)## [1] 0.1764706Answer: The probability of a customer is male and lives in ‘Other’ or is female and lives in ’Other is about 17.65%.
Q3) Two cards are drawn without replacement from a standard deck of 52 playing cards. What is the probability of choosing a diamond for the second card drawn, if the first card, drawn without replacement, was a diamond? Express your answer as a decimal number only.
# Total number of cards in a deck
total = 52
#Total number of card for 2nd draw (without replacement)
total = total - 1
# Total number of diamonds in a deck
D <- 13
# Total number of diamonds in a deck during 2nd draw
D <- D - 1
print(D)## [1] 12print(total)## [1] 51# Probability
prob <- (D / total)
print(prob)## [1] 0.2352941Answer: If the first draw is diamond and without replacement, the probability of drawing diamond again in the 2nd time is 23.52%.
Q4) A coordinator will select 10 songs from a list of 20 songs to compose an event’s musical entertainment lineup. How many different lineups are possible?
# Building a combination function
perm_without_replace <- function(n, r){
return(factorial(n)/factorial(n - r))
}
perm_without_replace(20,10)## [1] 670442572800Answer: The amount of lineups that are possible are 670,442,572,800.
Q5) You are ordering a new home theater system that consists of a TV, surround sound system, and DVD player. You can choose from 20 different TVs, 20 types of surround sound systems, and 18 types of DVD players. How many different home theater systems can you build?
# Variable values
tv <- 20
dvd <- 20
ss <- 18
# Number of ways to select
total <- (tv * dvd * ss)
print(total)## [1] 7200Answer: The total number of ways to select the home threater system is 670,442,572,800.
Q6) A doctor visits her patients during morning rounds. In how many ways can the doctor visit 10 patients during the morning rounds?
# Number of ways to visit 10 patients
factorial(10)## [1] 3628800Answer: There are, in total, 3,628,800 ways to visit the 10 patients.
Q7) If a coin is tossed 7 times, and then a standard six-sided die is rolled 3 times, and finally a group of four cards are drawn from a standard deck of 52 cards without replacement, how many different outcomes are possible?
# Number of ways of each event
coin <- 2^7
dice <- 6^3
## For cards, permutation without repetition
card <- (perm_without_replace(52,4))
# Total number of all possible outcomes of three events
outcome <- (coin * dice * card)
print(outcome)## [1] 179640115200Answer: The total possible outcomes for these three events are 179,640,115,200.
Q8) In how many ways may a party of four women and four men be seated at a round table if the women and men are to occupy alternate seats.
# Number of each sex
n = 4
## Here, we assume a man gets to be seated first
# Ways men can be seated
men <- factorial(n-1)
# Ways women can be seated
women <- factorial(n)
# Total ways possible
total <- (men * women)
print(total)## [1] 144Answer: The total number of ways to seat these people alternatively are 144.
Q9) An opioid urinalysis test is 95% sensitive for a 30-day period, meaning that if a person has actually used opioids within 30 days, they will test positive 95% of the time P( + | User) =.95. The same test is 99% specific, meaning that if they did not use opioids within 30 days, they will test negative P( - | Not User) = .99. Assume that 3% of the population are users. Then what is the probability that a person who tests positive is actually a user P(User | +)?
# Variable values
op <- 0.95 # (+|User) Positive positive rate
on <- 1 - op # (+|Non User) false positive rate
Upop <- 0.03 # Users in population
NUpop <- 1 - Upop # Non-users in population
# Bayesian probability
Userispos <- (Upop * op) / ((Upop * op) + (NUpop * on))
print(Userispos)## [1] 0.3701299Answer: The probability of having correctly decting a user that is positive is 37.01% based on bayesian probability.
Q10) You have a hat in which there are three pancakes. One is golden on both sides, one is brown on both sides, and one is golden on one side and brown on the other. You withdraw one pancake and see that one side is brown. What is the probability that the other side is brown? Explain.
# Probability of select one of the three pancakes
P <- 1/3
# Probability of selecting a hat that is brown, given the one side is brown and there are two pancakes that have at least one brown side
B <- 1/2
# P{brown, brown}
Pbb <- (P*B)
print(Pbb)## [1] 0.1666667Answer: The total probability of the other side of the hat drawn is brown is 16.67%.