DataAnalysisHW2

Problem Set

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. Hint: use expand.grid. rowSums, and length functions.

Each roll can result in a value from 1 to 6, and by using expand grid we can create a quick table of all potential totals after 3 rolls
```
##use expand.grid() to create a table with all possible 3-roll combos
grid<-expand.grid(roll1=1:6,roll2=1:6,roll3=1:6)

#sum up each combo
sums<-rowSums(grid)

#count how many sums are 12s
twelves<-length(which(sums==12))

#divide number of twelves by total combinations (6*6*6=216)
prob<-twelves/216

#Print result
prob
```
```
## [1] 0.1157407
```
This gives us a probability of 0.1157, or 11.57%
A newspaper company classifies its customers by gender and location of residence. The research department has gathered data from a random sample of customers. The data is summarized in the table below

What is the probability that a customer is male and lives in ‘Other’ or is female and lives in ‘Other’? Express your answer as a decimal number only. Show your R code.

Probability that a customer is male and lives in ‘Other’: 0.1176

Probability that a customer is male and lives in ‘Other’: 0.0588

Probability that a customer is male or female and lives in ‘Other’: 0.1765

Hint: create the matrix above in R, use rowSums and col.Sums to generate marginal probabilities.
```
#create matrix

males<-c(200,200,100,200,200)
females<-c(300,100,200,100,100)

customers<-matrix(c(males,females),nrow=5,byrow=FALSE)
customers
```
```
##      [,1] [,2]
## [1,]  200  300
## [2,]  200  100
## [3,]  100  200
## [4,]  200  100
## [5,]  200  100
```
```
#Get total customers
totalcustomers<-sum(colSums(customers))

male_other<-customers[5,1]
female_other<-customers[5,2]

prob_m_other<-(male_other/totalcustomers)
prob_m_other
```
```
## [1] 0.1176471
```
```
prob_f_other<-(female_other/totalcustomers)
prob_f_other
```
```
## [1] 0.05882353
```
```
prob_other<-(male_other+female_other)/totalcustomers
prob_other
```
```
## [1] 0.1764706
```
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. Show your R code.

After the first draw, there will be 11 diamonds and 51 total cards left, resulting in a probability of 0.2157
```
## Number of cards left after first draw 
cards_left<-52-1

##N Number of diamonds left
diamonds_left<-12-1

PDraw2_Diamond<-(diamonds_left/cards_left)
PDraw2_Diamond
```
```
## [1] 0.2156863
```

The next few questions are about permutation, combination, and factorials.

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? Show your R code.

There are 184,756 possible lineups of 10 songs.
```
lineups<-choose(20,10)
lineups
```
```
## [1] 184756
```
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? Show your R code

You can create 7200 different combinations with these options.
```
#This will just be a simple multiplication of factors
#Create all factors

tvs<-20
surroundsounds<-20
dvds<-18

combos<-tvs*surroundsounds*dvds
combos
```
```
## [1] 7200
```
A doctor visits her patients during morning rounds. In how many ways can the doctor visit 10 patients during the morning rounds? Show your R code

With each visit, the options for her next visit decrease by 1, making this a factorial case.

10! = 3,628,800 different ways she can order her morning rounds.
```
#Calculatie 10!
diffways<-factorial(10)

diffways
```
```
## [1] 3628800
```
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? Show your R code

coin flip outcomes: 128

dice roll outcomes: 216

card draw outcomes: 270,725

Different outcomes possible: 7,485,004,800
```
#calculate coinflip outcomes
coinflip_outcomes<-2^7
coinflip_outcomes
```
```
## [1] 128
```
```
#calculate dice roll outcomes
dice_outcomes<-6^3
dice_outcomes
```
```
## [1] 216
```
```
#calculate card draw outcomes
draw_outcomes<-choose(52,4)
draw_outcomes
```
```
## [1] 270725
```
```
total_outcomes<- coinflip_outcomes*dice_outcomes*draw_outcomes
total_outcomes
```
```
## [1] 7485004800
```
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. Show your R code

Because it is a round table we can just start with one man and see the permutations of the order of people to their right. 3!*4!=144
```
# 1 man is set, so 3 remain, along with the 4 women. 

male_order<-factorial(3)
female_order<-factorial(4)

seating_order<- male_order*female_order
seating_order
```
```
## [1] 144
```
BAYESIAN PROBABILITY 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 | +)?

P(A|B)=( P(B|A) * P(A) ) / P(B)
P(User | +) = ( P(+ | User) * P(User)) / P(+)

P(+ | User)= 0.95
P(User)= 0.03
P(+)= P(+ | User)*P(User) + P(+ | Not User) * P(Not User)
= 0.95*0.03 + 0.01*0.97
= 0.0382
```
# Given probabilities
p_user <- 0.03
p_not_user <- 1 - p_user
p_pos_given_user <- 0.95
p_pos_given_not_user <- 1 - 0.99  # specificity is 99%

# Total probability of testing positive
p_pos <- p_pos_given_user * p_user + p_pos_given_not_user * p_not_user

# Bayes' Theorem: P(User | +)
p_user_given_pos <- (p_pos_given_user * p_user) / p_pos

# Print the result
p_user_given_pos
```
```
## [1] 0.7460733
```
Probability that a person who tests positive is actually a user: 0.7460733 = 0.7461
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.

Let’s call these 3 pancakes Brown-Brown, Brown-Gold, and Gold-Gold. We withdraw one pancake and see a brown side. We know that means that we are either holding Brown-Brown or Brown-Gold. There are 3 sides that we could possibly be looking at and therefore 3 results for the remaining side:

Brown-Brown Side 1 (Remaining side = Brown)

Brown-Brown Side 2 (Remaining side = Brown)

Brown-Gold Side 1 (Remaining side = Gold)

2 of the 3 possible scenarios would result in the other side being brown. 2/3 = 0.6667

# Three pancakes: BB, GB, GG

total_brown<-3

#Possible favorable outcomes both come from BB
favorable<-2

prob_2ndBrown<-favorable/total_brown

prob_2ndBrown

## [1] 0.6666667

DataAnalysisHW2

Grant Cosovich

2025-09-07

Problem Set