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 consider a specific example, where \(n=5\) and \(k=10\).

With these numbers then \(P(Y=1)=\frac{10^5-9^5}{10^5}=0.4095\) There are \(10^5\) possible outcomes of the X distributions and \(10^5-9^5\) of them contain at least a single 1.

\(P(Y=2)=\frac{10^5-8^5-[10^5-9^5]}{10^5}=0.2681\) The numerator contains 3 parts. There are \(10^5\) possible outcomes, \(10^5-9^5\) contain at least a single 1 (calculated above) and \(10^5-8^5\) containing a 1 or a 2.

Generalizing with \(y=2\), the numerator can be written as \(k^n-(k-y)^n-[k^n-(k-y+1)^n]\) which simplifies to \(-(k-y)^n + (k-y+1)^n\)

Putting it all together

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

Your organization owns a copier or MRI. This machine has a manufacturer’s expected lifetime of 10 years. This means that we expect one failure every ten years.

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

The probability of breaking in a given year is \(P(S) = 0.1\) and the probability of not breaking is \(P(F)=1-P(S)=0.9\)

\(P(X > 8) = 1-P(X \leq 8) = 1-(1-0.9^8)=0.4305\)

1-(1-0.9**8)
## [1] 0.4304672

The expected value is \(\mu=E(X)=\frac{1}{p}=\frac{1}{.1}=10\)

The standard deviation is \(\sigma=\sqrt{\frac{1-p}{p^2}}=\sqrt{\frac{.9}{.1^2}}=9.4868\)

sqrt(.9/.1**2)
## [1] 9.486833

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

\(P(X > 8) = 1-P(X \leq 8)=1-(1-e^{-\lambda x}) = e^{-8(.1)}=0.4493\)

exp(-8*.1)
## [1] 0.449329

The expected value is \(\mu=E(X)=\frac{1}{\lambda}=\frac{1}{.1}=10\)

The standard deviation is \(\sigma=\sqrt{\frac{1}{\lambda^2}}=\sqrt{\frac{1}{.1^2}}=10\)

sqrt(1/.1**2)
## [1] 10

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

\(P(X > 8) = 1-P(X \leq 8)=1-P(X=0)+P(X=1)+P(X=2)...P(X=8)=0.5695\)

1-pbinom(0, 8, 0.1)
## [1] 0.5695328

The expected value is \(\mu=E(x)=8(.1)=.8\)

The standard deviation sis \(\sigma=np(1-p)=8(0.1)(0.9)=.72\)

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

\(P(x > 8;10)=1-P(x \leq 8; 10)=1-P(0;10)+P(1;10)+P(2;10)...+P(8;10)=0.6672\)

1-ppois(8, 10)
## [1] 0.6671803

The expected value is \(\mu=E(X)=\lambda=10\)

The standard deviation is \(\sigma=\sqrt{\lambda}=\sqrt{10}=3.1623\)