Question 1

Q1 :- 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.

Answer :- In the question it is stated that Y denotes the minimum of the Xis. Lets assume each independent random variable Xi has k possibilities.

Now each Xi has k possibilities: 1, 2, …__, k. Then total possible number of assignments for the entire collection of random variables are X1, X2, …, Xn is (k)^n . This will be denominator for probability distribution function.

The number of ways of getting Y = 1 is k^n - (k - 1)^n / k^n, since k^n represents the total number of options and (k-1)^n represents all of the options where none of the Xi’s are equal to 1.

When X = 1: P(X=1) = k^n - (k-1)^n / k^n

Similarly When X= 2 & 3: P(X=2) = (k-2+1)^n - (k-2)^n / k^n P(X=3) = (k-3+1)^n - (k-3)^n / k^n

Generalization this for (X=m): P(X=m) = (k-m+1)^n - (k-m)^n * k^n

Question 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.)

  1. 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..)
prob_fail <- 1/10 
n = 8
prob_notfail <- 1-prob_fail

prob_geom  <- 1-pgeom(n-1, prob_fail) 
prob_geom
## [1] 0.4304672
#Expected value 
expec_val <- 1/prob_fail
expec_val
## [1] 10
#Standard Deviation
SD <- sqrt(prob_notfail/(prob_fail^2))
SD 
## [1] 9.486833
  1. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.
n <- 8 
lambda <- 1/10

p_expo <- pexp(n, lambda, lower.tail=FALSE)
round(p_expo,2)
## [1] 0.45
#Expected value = 1/lambda
expec_val <- 1/lambda
expec_val
## [1] 10
#SD = 1/??^2
SD <- sqrt(1/lambda^2)
SD 
## [1] 10
  1. :- 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) (n Choose k) p^k * (1-p)^(n???k)
n <- 8
p <- 1/10
q <- (1-p)
k <- 0


p_binomial <- dbinom(k, n, p)
p_binomial
## [1] 0.4304672
#Expected value
exp_val <- n * p
exp_val
## [1] 0.8
# Standard deviation
sd_bino <- sqrt(n*p*q) 
sd_bino
## [1] 0.8485281
  1. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.
#Since average number of failures in every 10 years is 1 so average number of failures in 8 years will be:

lambda <- 8/10
k <- 0

ppois(0,lambda = .8 )
## [1] 0.449329
#Expected value 
exp_val <- 8/10
exp_val
## [1] 0.8
# Standard deviation
SD <- sqrt(8/10)
SD
## [1] 0.8944272