Problem 1

Let X1,X2,…,Xn be n mutually independent random variables, each of which is uniformly distributed on the integers from 1 to k. Let Y denote the minimum of the Xi’s. Find the distribution of Y.

\(\mathrm{Y}\) is the probability distribution for a given value \(a(1 \leq a \leq k)\) and \(Y\) will be greater than or equal to \(a\) if and only if \(X 1, X 2, \ldots, X n\) are greater than or equal to a. \[ P(Y(1 \leq a \leq k))=\frac{(k-a+1)^n-(k-a)^n}{k^n} \] The probability that \(X 1\) is greater than or equal to \(a\) is \((k-y+1) / k\) : \[ P(X 1 \geq a)=\frac{(k-a+1)^n}{k^n} \] because there are \((k-a+1)\) possible integers from \(a\) to \(k\). Similarly, the probability that \(\mathrm{X}_2, \mathrm{X}_3, \ldots, \mathrm{Xn}\) are each greater than or equal to \(\mathrm{y}\) is also ( \(\mathrm{k}\) \(-y+1) / k\), and since these random variables are independent, we can multiply their probabilities. So, the probability that \(Y\) is greater than or equal to \(a\) is: \[ P(Y \geq a)=P(X 1 \geq a){ }^* P(X 2 \geq a){ }^* . .{ }^* P(X n \geq a)=((k-y+1) / k)^{\wedge} n \] To find the probability that \(Y\) equals \(a\), we can calculate the difference between the probabilities of \(Y\) being greater than or equal to \(a\) and \(Y\) being greater than or equal to \((a+1)\) : \[ \begin{gathered} P(Y=a)=P(Y \geq a)-P(Y \geq a+1) \\ =\frac{(k-a+1)^n}{k^n}-\frac{(k-(a+1)+1)^n}{k^n} \\ =\frac{(k-a+1)^n}{k^n}-\frac{(k-a)^n}{k^n} \end{gathered} \]

# Function to calculate the probability that Y is equal to a
probability_Y_equals_a <- function(a, n, k) {
  prob_Y_greater_than_a <- ((k - a + 1)^n - (k - a)^n) / (k^n)
  return(prob_Y_greater_than_a)
}

# Parameters
n <- 5  
k <- 20 

# Distribution of Y for each 'a'
distribution_Y <- sapply(1:k, function(a) probability_Y_equals_a(a, n, k))

# A data frame to represent the distribution
distribution_df <- data.frame(Y = 1:k, Probability = distribution_Y)
print(distribution_df)
##     Y  Probability
## 1   1 0.2262190625
## 2   2 0.1832909375
## 3   3 0.1467846875
## 4   4 0.1160253125
## 5   5 0.0903753125
## 6   6 0.0692346875
## 7   7 0.0520409375
## 8   8 0.0382690625
## 9   9 0.0274315625
## 10 10 0.0190784375
## 11 11 0.0127971875
## 12 12 0.0082128125
## 13 13 0.0049878125
## 14 14 0.0028221875
## 15 15 0.0014534375
## 16 16 0.0006565625
## 17 17 0.0002440625
## 18 18 0.0000659375
## 19 19 0.0000096875
## 20 20 0.0000003125
probability_Y_equals_a <- function(k,n,trials=100000) {
  Y<-rep(0,trials)
  for (i in 1:trials) {
    x<-sample.int(k,size=n,replace=TRUE)
    Y[i]<-min(x)
  }
  return(Y)
}

par(mfrow=c(1,2))
hist(probability_Y_equals_a(k,n),breaks=60,main=paste("Simulation with k=",k," and n=",n,sep=""),xlab="Y",xlim=c(1,k))
pY<-((k-1:k+1)^n-(k-1:k)^n)/k^n
barplot(pY,main=paste("Theoretical with k=",k," and n=",n,sep=""),xlab="Y",xlim=c(1,k))

The ‘sample.int’ function generates random samples from 1 to k with replacement to simulate the random variables X1, X2, …, Xn, and then calculates the minimum Y for each set of random variables. The histograms show the frequency or probability of different values of Y.nd compare the simulated and theoretical results, helping you understand how the minimum of n random variables is distributed.

Problem 2

Your organization owns a copier (future lawyers, etc.) or MRI (future doctors). This machine has a manufacturer’s expected lifetime of 10 years. This means that we expect one failure every ten years. (Include the probability statements and R Code for each part.).

Problem 2 (a)

What is the probability that the machine will fail after 8 years? Also provide the expected value and standard deviation. Model as a geometric. (Hint: the probability is equivalent to not failing during the first 8 years.)

Let a geometric random \(X\) with parameter \(P\) is the probability that the machine will fail after 8 years is: - \(P(\) Machine fails in the first 8 years)) \[ P(X>\mathrm{k})=1-P(X \leq \mathrm{k})=1-\left(1-p^{k+1}\right)=p^{k+1} \] \(\mathrm{k}\) is he number of failures before the first success and p is probability of success.

For \(\mathrm{k}=8\), \[ P(X>8)=0.9^9 \]

# Probability of  one failure in every ten years
p_failure <- 1/10

# Probability that the machine will fail after 8 years (P(X > 8))
p_failure_after_8_years <- pgeom(8, p_failure, lower.tail = FALSE)
cat("Probability that the machine will fail after 8 years is: ", p_failure_after_8_years, "\n")
## Probability that the machine will fail after 8 years is:  0.3874205
# Expected value (mean)
expected_value <- (1 - p_failure) / p_failure

# Standard deviation 
standard_deviation <- sqrt((1 - p_failure) / (p_failure^2))

cat("Expected Value (Mean): ", expected_value, "\n")
## Expected Value (Mean):  9
cat("Standard Deviation σ is: ", standard_deviation, "\n")
## Standard Deviation σ is:  9.486833

Problem 2(b)

What is the probability that the machine will fail after 8 years? Also provide the expected value and standard deviation. Model as an exponential?

The probability that the machine will fail after 8 years, modeled as an exponential distribution, and using the cumulative distribution function (CDF) of the exponential distribution. The CDF for the exponential distribution is given by: \[ P(X \leq k)=1-e^{-\lambda k} \] where \(P(X \leq k)\) is the probability that the machine fails within the first \(k\) years \(k\) is the time ( 8 years) \(\lambda\) is the failure rate of the exponential distribution, \(\lambda=0.1\) To calculate the probability that the machine fails after 8 years \((P(X>8))\), using the complementary probability: \[ P(X>8)=1-P(X \leq 8)=1-\left(1-e^{-\lambda \cdot 8}\right) \]

lambda <- 0.1
k <- 8
# Probability that the machine fails within the first 8 years (P(X <= 8))
p_failure_within_8_years <- 1 - exp(-lambda * k)

# Probability that the machine fails after 8 years (P(X > 8))
p_failure_after_8_years <- 1 - p_failure_within_8_years
cat("Probability that the machine will fail after 8 years is: ", p_failure_after_8_years, "\n")
## Probability that the machine will fail after 8 years is:  0.449329

Expected Value (Mean) for an Exponential Distribution: \(E(X)=\frac{1}{\lambda}=\frac{1}{0.1}=10\)

Standard Deviation for an Exponential Distribution: $ σ = =10$

Problem 2(c)

What is the probability that the machine will fail after 8 years? Also provide the expected value and standard deviation. Model as a binomial. (Hint: 0 success in 8 years)

Probability of \(k\) successes in \(n\) trials with probability of success \(p\) : \[ P(X=k)=\left(\begin{array}{l} n \\ k \end{array}\right) \cdot p^k \cdot(1-p)^{n-k} \] \(\mathrm{X}\) is the number of times the machine fails in \(\mathrm{n}=8\) years. Expected Value (Mean) of a binomial distribution: \[ E(X)=n \cdot p \] Standard Deviation of a binomial distribution: \[ S D(X)=\sqrt{n \cdot p \cdot(1-p)} \]

p_failure <- 1/10
p_success <- 1 - p_failure
n <- 8

# Probability of 0 failures in 8 years using the binomial distribution
prob_0_failures_in_8_years <- pbinom(p_success, n, p_failure)

# Expected value (mean) of the number of failures in 8 years
expected_value <- n * p_failure

# Standard deviation of the number of failures in 8 years
standard_deviation <- sqrt(n * p_success * p_failure)

# Print the results
cat("Probability that the machine will fail after 8 years: ", prob_0_failures_in_8_years, "\n")
## Probability that the machine will fail after 8 years:  0.4304672
cat("Expected Value (Mean) of failures in 8 years: ", expected_value, "\n")
## Expected Value (Mean) of failures in 8 years:  0.8
cat("Standard Deviation of failures in 8 years: ", standard_deviation, "\n")
## Standard Deviation of failures in 8 years:  0.8485281

Problem 2 (D)

What is the probability that the machine will fail after 8 years? Also provide the expected value and standard deviation. Model as a Poisson.

Probability Mass Function (PMF) for Poisson Distribution: \[ P(X=k)=\frac{e^{-\lambda} \cdot \lambda^k}{k !} \] Where: - \(\quad X\) represents the number of failures in a fixed interval. - \(\quad k\) is the number of failures of interest - \(\quad \lambda\) is the average rate of failures in the fixed interval. Expected Value (Mean) of a Poisson Distribution: \[ E(X)=\lambda \] Standard Deviation of a Poisson Distribution: \[ \sigma(X)=\sqrt{\lambda} \]

lambda <- 8 / 10
# Number of failures of interest (0 failures in 8 years)
k <- 0

# Probability of 0 failures in 8 years using the Poisson distribution
prob_0 <- ppois(k, lambda)

# Expected Value (Mean) of the number of failures in 8 years
expected_value <- lambda

# Standard Deviation of the number of failures in 8 years
standard_deviation <- sqrt(lambda)

# Print the results
cat("Probability that the machine will fail after 8 years (0 failures in 8 years): ", prob_0, "\n")
## Probability that the machine will fail after 8 years (0 failures in 8 years):  0.449329
cat("Expected Value (Mean) of failures in 8 years: ", expected_value, "\n")
## Expected Value (Mean) of failures in 8 years:  0.8
cat("Standard Deviation of failures in 8 years σ is: ", standard_deviation, "\n")
## Standard Deviation of failures in 8 years σ is:  0.8944272