Question 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 .

Y is the minimum of the Xi’s

So, P(Y > y) = P(X1 > y and X2 > y ……..Xn > y)

Using multiplication rule we can write below for independent Xi’s

P(Y > y) = P(X1 > y) × P(X2 > y) × … × P(Xn > y)

We know that Xi’s are uniformly distributed on {1, 2, …, k} so we can write:

P(Xi > y) = (k - y) / k

Therefore, we can write:

P(Y > y) = [(k - y) / k]^n

y = 1, 2, …, k.

The probability that Y takes on the value y is then:

P(Y = y) = P(Y > y-1) - P(Y > y) = [(k - (y-1)) / k]^n - [(k - y) / k]^n

y = 1, 2, …, k.

So Y is a discrete random variable that takes on values in {1, 2, …, k}.

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.).

Question 2a:

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..)

Formula for geometric distribution: \[PP(X=k) = (1-p)^{k-1} \cdot p\]

where:

  • P(X = k) is the probability that the first success occurs on the kth trial.
  • (1 - p)^(k-1) is the probability of having k-1 failures before the first success.
  • p is the probability of success on each trial.

Expected value:

E(X) = 1/p

Var(X) = (1-p)/p^2.

The probability that the machine will fail after 8 years is equivalent to the probability of not failing during the first 8 years. So, we need to find P(X > 8).

P(X > 8) = P(X = 9) + P(X = 10)

The probability of the machine failing in the 9th year is:

P(X = 9) = (1-p)^8 * p

The probability of the machine failing in the 10th year is:

P(X = 10) = (1-p)^9 * p

Therefore,

P(X > 8) = P(X = 9) + P(X = 10) = (1-p)^8 * p + (1-p)^9 * p = (9/10)^8 * (1/10) + (9/10)^9 * (1/10) = 0.4305

So, the probability that the machine will fail after 8 years is 0.4305.

Expected Value:

E(X) = 1/p = 1/(1/10) = 10

Variance:

Var(X) = (1-p)/p^2 Standard deviation:

SD(X) = sqrt(Var(X)) = sqrt((1-p)/p^2) = 3.1623

R code:

p <- 1/10
prob_not_failing_8_years <- dgeom(8, p, log = FALSE)
prob_not_failing_8_years 
## [1] 0.04304672
# Expected value 
expected_value <- mean(rgeom(1000000, p))
expected_value 
## [1] 9.000645
# Standard deviation 
standard_deviation <- sd(rgeom(1000000, p))
standard_deviation  
## [1] 9.497055

Question 2b:

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

cumulative distribution function of the exponential distribution:

Lamda = 1/10

P(X > 8) = 1 - F(8) = 1 - (1 - e^(-lambda * 8)) = e^(-lambda * 8) = e^(-8/10) = 0.4493

where X is the time until failure and F(x) is the cumulative distribution function (CDF) of the exponential distribution.

Expected value:

E(X) = 1 / lambda

In this case, the expected value is:

E(X) = 1 / (1/10) = 10

The standard deviation of an exponential distribution is also given by:

SD(X) = 1 / lambda

In this case, the standard deviation is:

SD(X) = 1 / (1/10) = 10

R code:

# Probability of not failing during the first 8 years
p <- 1/10
prob_not_failing_8_years <- pexp(8, rate = 1/p, lower.tail = FALSE)
prob_not_failing_8_years  
## [1] 1.804851e-35
# Expected value
expected_value <- 1/p
expected_value  
## [1] 10
# Standard deviation
standard_deviation <- 1/p
standard_deviation 
## [1] 10

Question 2c:

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)

Binomial formula; \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]

where

\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]

\[p^k (1-p)^{n-k}\]

is the probability of having exactly k successes and (n - k) failures in n trials.

Expected value:

E(X) = n * p

Standard deviation:

SD(X) = sqrt(n * p * (1 - p))

Solve using binomial

The probability of having 0 failures in 8 years is:

P(X = 0) = (8 choose 0) * (1/10)^0 * (9/10)^8 = 1 * 1 * 0.4304672 = 0.4304672

where:

  • (8 choose 0) = 1 is the number of ways to choose 0 failures out of 8 years

  • (1/10)^0 = 1 is the probability of a failure in one year

  • (9/10)^8 = 0.4304672 is the probability of not having a failure in 8 years.

Expected value:

E(X) = n * p = 8 * (1/10) = 0.8

Standard deviation:

SD(X) = sqrt(n * p * (1 - p)) = sqrt(8 * (1/10) * (9/10)) = 0.7483315

Therefore, the probability that the machine will fail after 8 years is 0.4305, the expected value is 0.8 failures in 8 years, and the standard deviation is 0.7483 failures in 8 years.

Probability of not failing during the first 8 years

R code:

p <- 1/10
prob_not_failing_8_years <- dbinom(0, size = 8, prob = p)
prob_not_failing_8_years 
## [1] 0.4304672
Expected value of the binomial distribution
expected_value <- 8 * p
expected_value  
## [1] 0.8
Standard deviation of the binomial distribution
standard_deviation <- sqrt(8 * p * (1 - p))
standard_deviation 
## [1] 0.8485281

Question 2d:

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

The formula for the Poisson distribution:

\[P(x) = \frac{e^{-\lambda}\lambda^x}{x!}\]

Where:

  • P(x) is the probability of observing x events in a fixed interval of time or space
  • e is the mathematical constant approximately equal to 2.71828
  • λ is the expected number of events in the interval
  • x is the actual number of events observed in the interval
  • x! is the factorial of x, which is the product of all positive integers up to x.
Probability of the machine failing after 8 years

R code:

lamda<-0.8
e<-2.71828
prob_fail_8years <- dpois(1, lambda = 0.8)
prob_fail_8years
## [1] 0.3594632
# Expected value of the number of failures in 8 years
expected_fail_8years <- 0.8
expected_fail_8years 
## [1] 0.8
Standard deviation of the number of failures in 8 years
sd_fail_8years <- sqrt(0.8)
sd_fail_8years 
## [1] 0.8944272