Week 6 - Probability

library(tidyverse)

## Warning: package 'ggplot2' was built under R version 4.3.2

library(openintro)

Question 1

A bag contains 5 green and 7 red jellybeans. How many ways can 5 jellybeans be withdrawn from the bag so that the number of green ones withdrawn will be less than 2?

# Calculate combinations when 0 or 1 green jellybeans are chosen
green0 <- factorial(7) / (factorial(5) * factorial(7 - 5))
green1 <- (factorial(5) / (factorial(1) * factorial(5 - 1))) * (factorial(7) / (factorial(4) * factorial(7 - 4)))

# Total combinations
green0 + green1

## [1] 196

Question 2

A certain congressional committee consists of 14 senators and 13 representatives. How many ways can a subcommittee of 5 be formed if at least 4 of the members must be representatives?

# Function to calculate combinations
nCr <- function(n, r) {
  factorial(n) / (factorial(r) * factorial(n - r))
}


senators <- 14
representatives <- 13

# Total combinations
nCr(representatives, 5) + nCr(representatives, 4) * nCr(senators, 1)

## [1] 11297

Question 3

# Calculate total outcomes
coin <- 2^5
die  <- 6^2
card <- 52 * 51 * 50

coin * die * card

## [1] 152755200

Question 4

# Find the probability and round it to 4 decimals
round(1 - (48/52) * (47/51) * (46/50), 4)

## [1] 0.2174

Question 5

# Function to calculate combinations
nCr <- function(n, r) {
  factorial(n) / (factorial(r) * factorial(n - r))
}


# Calculate combinations of choosing 5 movies from total
a <- nCr(17 + 14, 5)

# Calculate number of combinations with at least one mystery
b <- nCr(17 + 14, 5) - nCr(17, 5)

print(a)

## [1] 169911

print(b)

## [1] 163723

Question 6

# Function to calculate combinations
nCr <- function(n, r) {
  factorial(n) / (factorial(r) * factorial(n - r))
}


# Output the result in scientific notation rounded to the hundredths place
format(nCr(4, 3) * nCr(104, 3) * nCr(17, 3), scientific = TRUE, digits = 3)

## [1] "4.95e+08"

Question 7

# Function to calculate permutations
nP <- function(n, r) {
  factorial(n) / factorial(n - r)
}

total_permutations <- 0

# Iterate through possible combinations of books
for (novel_count in 0:min(6, 13)) {
  for (play_count in 0:min(6, 13 - novel_count)) {
    for (poetry_count in 0:min(7, 13 - novel_count - play_count)) {
      nonfiction_count <- 13 - novel_count - play_count - poetry_count
      if (nonfiction_count <= 4) {
        permutations <- nP(6, novel_count) * nP(6, play_count) * nP(7, poetry_count) * nP(5, nonfiction_count)
        total_permutations <- total_permutations + permutations
      }
    }
  }
}

#  Step 1
print(format(total_permutations, scientific = TRUE, digits = 3))

## [1] "3e+10"

# Step 2
remaining <- nP(6 + 7 + 5, 13 - 6)

# Output the result for step 2
print(format(remaining, scientific = TRUE, digits = 2))

## [1] "1.6e+08"

Question 8

2 / factorial(10)

## [1] 5.511464e-07

Question 9

# Calculate expected value 1
expected_value1 <- round((44/52)*4+(8/52)*(-16),2)

# Calculate expected value 2
expected_value2 <- round((44/52)*833*4 + (8/52)*833*(-16),2)

print(expected_value1)

## [1] 0.92

print(expected_value2)

## [1] 768.92