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 .

Solution 1
Assuming that each \(X_{i}\) has k possibilities, we will have \(k^{n}\) possible values in total. 1 is the minimum of the Xi’s, the number of ways of getting Y.
for \(P(x = 1) = \frac{k^{n} - (k-1)^{n}}{k^{n}}\)
for \(P(x = 2) = \frac{(k-2+1)^{n} - (k-2)^{n}}{k^{n}}\)
For \(x_{1}, x_{2}, . . . , x_{n}\) mutually independent random variables, If Y = j then, the minimum value j is given as:
\(P(x = j) = \frac{(k-j+1)^{n} - (k-j)^{n}}{k^{n}}\)

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 2a (Geometric)
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..)
Solution 2a (Geometric)
Probability of failure in 8 years
Given that p is the probability of success and q = 1 - p is the probability of failure for one trial, the geometric probability distribution is given by: \(P(X=x) = p(1−p)^{x-1}\), x = 1, 2, 3, … where x is the number of trials up to the first success

x = 8
p = 1 / 10 # probability of success
prob_geometric = p * ((1 - p)^(8 - 1))
prob_fail_8years_geometric =  round(prob_geometric, 4)
prob_fail_8years_geometric
## [1] 0.0478

Expected value
The expected value of a geometric distribution is given by \(E(X) = 1/p\)

expected_value_geometric = 1/p
expected_value_geometric
## [1] 10

Standard Deviation
The standard deviation of a geometric distribution is given by \(S.D = \frac{\sqrt{1-p}}{p}\)

sd_geom = sqrt(1 - p)/p 
sd_geometric = round(sd_geom, 4)
sd_geometric
## [1] 9.4868


Problem 2b (Exponential)
What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.
Solution 2b (Exponential)
Probability of failure in 8 years
Given the rate parameter \(\lambda\), The exponential distribution is given by:
\(P(X=x) = \lambda e^{-\lambda x}\)

rate = 1/10
x = 8
prob_exponetial = rate * exp(-rate * x)
prob_fail_8years_exponential = round(prob_exponetial, 4)
prob_fail_8years_exponential
## [1] 0.0449

Expected value
The expected value of an exponential distribution is given by \(E(X) = 1/\lambda\)

expected_value_exponential = 1/rate
expected_value_exponential
## [1] 10

Standard Deviation
The standard deviation of an exponential distribution is given by \(\sigma = \frac{1}{\lambda}\)

sd_expo = 1/rate
sd_exponential = round(sd_expo, 4)
sd_exponential
## [1] 10


Problem 2c (Binomial)
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)
Solution 2c (Binomial)
Probability of 0 success in 8 years
Given the probability of success p, the probability of x successes in n trials is given by:
\(P(x,n,p)=\binom{n}{x}p^{x}(1−p)^{n-x}\)

p = 1/10
prob_0_success = choose(8, 0) * p^0 * (1-p)^8
prob_0_success_in_8_years = round(prob_0_success, 4)
prob_0_success_in_8_years
## [1] 0.4305

OR

prob = dbinom(0, 8, p)
prob_0_success_in_8_trials = round(prob, 4)
prob_0_success_in_8_trials
## [1] 0.4305

Expected value
The expected value of a binomial distribution is given by \(E(X) = np\)

n = 8
p = 1/10
expected_value_binomial = n*p
expected_value_binomial
## [1] 0.8

Standard Deviation
The standard deviation of a binomial distribution is given by \(\sigma = \sqrt{npq}\)

sd_binom = sqrt(n*p*(1-p))
sd_binomial = round(sd_binom, 4)
sd_binomial
## [1] 0.8485


Problem 2d (Poison)
What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.
Solution 2d (Poison)
Probability of failure in 8 years
The poison distribution formula is given by:
\(P(X=x) = \frac{\lambda ^{x}}{x!}e^{-\lambda}\)
\(\lambda = 10\)

rate = 8 * (1 / 10)
e <- exp(1)
prob_poison = (rate^0 * exp(-rate)) / factorial(0)
prob_fail_8years_poison = round(prob_poison, 4)
prob_fail_8years_poison
## [1] 0.4493

OR

prob_pois = dpois(0, rate)
prob_poison_8years = round(prob_pois, 4)
prob_poison_8years
## [1] 0.4493

Expected value
The expected value of the poison distribution is given by: \(E(X) = \lambda\)

expected_value_poison <- rate
expected_value_poison
## [1] 0.8

Standard Deviation
The standard deviation of the poison distribution is given by:
\(\sigma = \sqrt{\lambda}\)

sd_pois = sqrt(rate)
sd_poison = round(sd_pois, 4)
sd_poison
## [1] 0.8944