1) Let \(X_{1}\), \(X_{2}\), . . . , \(X_{n}\) 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 \(X_{i}s\). Find the distribution of Y .
Let’s say k is the number of options and n is the number of values. Number of combinations for \(X_{i}s\) is \(k^{n}\).
Probability of number of combinations with at least one 1, which would be total number of combinations minus all combinations greater than 1.
P(Y=1) = (\((k^{n})\) - \((k-1)^{n}\)) / \(k^{n}\)
Probability of number of combinations that have atleast one 2, but do not have 1. i.e. subtracting the previous numerator
P(Y=2) = ( (\((k^{n})\) - \((k-2)^{n}\)) - [\((k^{n})\) - \((k-1)^{n}\)] ) / \(k^{n}\)
= ( - \((k-2)^{n}\)) + \((k-1)^{n}\) ) / \(k^{n}\)
= ( \((k-1)^{n}\) - \((k-2)^{n}\)) ) / \(k^{n}\)
Based on the pattern,
P(Y=i) = ( \((k- (i-1))^{n}\) - \((k-i)^{n}\)) ) / \(k^{n}\)
where i is the minimum value.
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.).
a. 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..)
Probability of machine failure is p = 1/10 and q = 1 - 1/10 = 9/10
Based on Geometric distribution modelling,
P(T=n) = \(q^(n-1)p\)
Probability that the machine not failing in first 8 years is
P(X>8) = 1- P(X<=8) P(X<=8) = 1 - (1 - P(X>8))
P(X<=k) = \(q^(k+1)\)
= \(0.9^(8+1)\)
= 0.3874
Standard deviation= \(\sqrt{(0.9/0.1) ^2}\) ~ 9 (approx.)
Using R
p <- 0.1
q <- 0.9
pgeom(8, p, lower.tail = FALSE)## [1] 0.3874205The probability that the machine will fail after 8 years is approx ).3874 with a standard deviation around 9.
b. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.
For exponential model, \(\lambda\) = 0.1 Based on exponential function:
P(X<=k) = \(1 - e^(-k/\mu)\) where \(\mu = 1/\lambda\) = \(1 - e^(-k\lambda)\)
P(X>8) = 1 - P(X<=8) = 1 - \((1 - e^(-8*0.1))\)
1 - (1 - exp(-0.8))## [1] 0.449329Expected value = \(1/\lambda\) = 1/0.1 = 10 Standard deviation = \(\sqrt{(1/0.1) ^2}\) = 10
Calculating using R
pexp(8, 0.1, lower.tail = FALSE)## [1] 0.449329c. 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)
Expected value = number of years(8) * failure probability (0.1) = 0.8.
Using binomial distribution, the chance of any number of outcomes is:
\(P(Fail)^(nFails)\) * \(P(noFail)^(nNotFails)\)
Probability of No failure within 8 years:
0.1^0 * (1-0.1)^8 = 0.43
Calculating using R:
year = 8
pbinom(0, 8, 0.1)## [1] 0.4304672d. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.
Using poisson distribution, lambda is number of years (8 years) * 0.1 = 0.8
Standard deviation is the sqrt of the variance
variance <- 0.8
(stdev <- sqrt(variance))## [1] 0.8944272failure_rate = 0.1 num_years = 8 \(\lambda = 0.1 * 8 = 0.8\)
ppois(0, 0.1)^8## [1] 0.449329Alternate approach:
Probability of not failing before 8 years: \((\lambda^k * e^-\lambda)/k!\)
lambda <- 0.8
k <- 0
((lambda**k) * exp(-lambda)) / factorial(k)## [1] 0.449329