\[P(Y = y)\]

\[ P(Y = y) = P(X_i = y \text ,X_i > y) \]

\[ P(X_i = y \text, X_i > y) = P(X_1 = y) \times P(X_2 > y) \times \ldots \times P(X_n > y) \]

\(P(X_i = y) = \frac{1}{k}\), $\(P(X_i > y) = \frac{k - y}{k}\)

\[ P(Y = y) = \left( \frac{1}{k} \right) \times \left( \frac{k - y}{k} \right)^{n - 1} \]

\[( y = 1, 2, …, k )\]

\[P(Y = y) = \left\{ \begin{array}{ll}0 & \text{if } y < 1 \\\left( \frac{1}{k} \right) \times \left( \frac{k - y}{k} \right)^{n - 1} & \text{if } 1 \leq y \leq k \\0 & \text{if } y > k \end{array} \right. \]

#  expected lifetime in years
lifetime <- 10

# failure rate (lambda)  one failure for every ten years
lambda <- 1 / lifetime

# Probability of failure in each year
PoF <- 1 - exp(-lambda)

cat("Probability of failure in each year:", PoF, "\n")
## Probability of failure in each year: 0.09516258

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

\[ P(X=k)=(1−p)^{k−1}\times p\]

\[ P(failafter8 ears)=1−(1−p)^{8−1} \]

# Probability of failure each year
PoF <- 1 - exp(-lambda)

# Probability of not failing in 8 years
PnoF <- (1 - PoF) ^ 8

# Probability of failing after 8 years
PoFa <- 1 - PnoF

cat("Probability of failing after 8 years:", PoFa, "\n")
## Probability of failing after 8 years: 0.550671
EV <- 1 / PoF

sd <- sqrt((1 - PoF) / PoF^2)

cat("Expected value:", EV, "\n")
## Expected value: 10.50833
cat("Standard deviation:", sd, "\n")
## Standard deviation: 9.995835

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

\[ P(fails after 8 years)=1−(1−e^ {−λ×8})\]

lambda <- 1 / 10  

# Probability that machine fails after 8 years 
Pfa8 <- 1 - pexp(8, lambda)

EV <- 1 / lambda

sd <- 1 / lambda

print(Pfa8)
## [1] 0.449329
print(EV)
## [1] 10
print(sd)
## [1] 10

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

\[P(X=k)=(\frac{n}{k})×p^{k} ×(1−p)^{n−k}\] \[P(fails after 8 years)=P(X=0)\]

period <- 8

# Probability of failure in each year
PoF <- 1/10

# Probability of 0 failures in 8 years using binomial distribution
PnF <- dbinom(0, size = period, prob = PoF)

# Print the probability of 0 failures in 8 years
cat("Probability of 0 failures in 8 years:", PnF, "\n")
## Probability of 0 failures in 8 years: 0.4304672
EV <- period * PoF

sd <- sqrt(period * PoF * (1 - PoF))

cat("Expected value:", EV, "\n")
## Expected value: 0.8
cat("Standard deviation:", sd, "\n")
## Standard deviation: 0.8485281

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

\[P(X=k)= \frac{e^{−λ}\times λ^k} {k!}\]

period <- 8

# Probability of failure in each year
prob_failure <- 1 - exp(-lambda)

# Expected number of failures in 8 years
expected_failures <- lambda * period

# Probability of the machine failing after 8 years 
PoFa <- 1 - ppois(0, lambda = expected_failures)

cat("Probability of failing after 8 years:", PoFa, "\n")
## Probability of failing after 8 years: 0.550671
EV <- expected_failures

sd <- sqrt(expected_failures)

cat("Expected value:", EV, "\n")
## Expected value: 0.8
cat("Standard deviation:", sd, "\n")
## Standard deviation: 0.8944272