The probabiblity that any single \(X_i\) is greater that y \(\frac {k-y}{k}\), so \(k-y\) is greater than y out of total of k.
$P(Y = y) = ()^n - ()^n $
for \(y = 1,2,...,k\).
Y is the minimum of \(X_1,X_2...X_n\) where \(X_i\) is uniformly distributed.
# Probability of Success
p <- 9/10
# Probability the mission will fail after 8 years
fail_prob <- 1 - p^8
cat("Probability the machine will fail after 8 years", fail_prob,"\n")
## Probability the machine will fail after 8 years 0.5695328
#Expected value
cat("The expected value is ", 1/p,"\n")
## The expected value is 1.111111
# Standard Deviation
cat("The Standard deviation is ", sqrt((1-p)/(p^2)),"\n")
## The Standard deviation is 0.3513642
# rate of lambda
lambda <- 1/10
cat("The probability that the machine will fail " , exp(-lambda * 8), "\n")
## The probability that the machine will fail 0.449329
cat("The expected value", 1/ lambda,"\n")
## The expected value 10
cat("The standar deviation is: ", 1/ lambda,"\n")
## The standar deviation is: 10
The probability that there is 0 failure after 8 years. \(P(X=8) = \sum_{0}^n\dbinom{n}{k}(1-p)^n-k\)
Asumming \(p = 1/10\) and \(n=8\)
# probability of success
p <- 1/10
#number of years
n <- 8
# the number of failure
# number of failures in first 8 years
failures <- 0
cat("Probaility that machine will face after 8 years",pbinom(failures,n,p),"\n")
## Probaility that machine will face after 8 years 0.4304672
Let calculate the probability that the machine will fail after 8 years. \(P(X=8) = \sum_{0}^n\dbinom{8}{0}\frac {\lambda^x_e-\lambda}{x!}\)
lambda <- 1
cat("The probability that the machine will fail after 8 years: ", 1 - ppois(7, lambda*8),"\n")
## The probability that the machine will fail after 8 years: 0.5470392
# Expected Value
cat("The Expected Value is: ", lambda * 8, "\n")
## The Expected Value is: 8
# Standard Deviation
cat("The Standard Deviation is ", sqrt(lambda * 8),"\n")
## The Standard Deviation is 2.828427