Problem 1

We can express the distribution of Y with 1 - the remainder of values in n

\[P(x \leq Y) = 1 - P(x > Y)\] Now lets find the function to describe P(x > Y).

\[P(x > Y) = (\frac{k-y}{k})^n\] This gives us

\[P(x \leq Y) = 1 - (\frac{k-y}{k})^n\]

Problem 2

Geometric

  1. Probability
p_fail <- 1/10
p_nfail <- 1 - p_fail

n <- 8
px = p_nfail^(n-1) * p_fail
px
## [1] 0.04782969

Expected Value

ev = 1/p_fail
ev
## [1] 10

Standard Deviation

sd = sqrt(p_nfail/(p_fail^2))
sd
## [1] 9.486833
  1. Exponential

Probability

mu <- 10
k <- 8

p <- exp(-k/mu)
p
## [1] 0.449329

Expected Value

lambda <- 1/10
ev <- 1/lambda
ev
## [1] 10

Standard Deviation

sd <- sqrt(1/lambda^2)
sd
## [1] 10
  1. Binomial Distibution

Probability

n <- 8
x <- 0
p_fail <- 1/10
p = p_fail^x * (1 - p_fail)^(n-x)

p
## [1] 0.4304672

Expected Value is 0.8 failures in the 8 year period

ev = n*p_fail
ev
## [1] 0.8

Standard Deviation

sd = sqrt(8 * p_fail * (1-p_fail))
sd
## [1] 0.8485281
  1. Poisson

Probability

n <- 8
p_fail <- 1/10
t <- 1

lambda <- (n*p_fail) / t
p <- lambda^n * exp(-lambda/n)
p
## [1] 0.1518065

Expected Value

ev <- lambda
ev
## [1] 0.8

Standard Deviation

sd <- sqrt(lambda)
sd
## [1] 0.8944272