I know I probably butchered this problem. I might have misunderstood what it was asking of me.
My interpretation:
Y is the minimum \(X\) variable, so \(1 \leq y \leq k\)
\(k\) = all the integers in the set from 1 to \(k\). The total number of possibilities that \(y\) is the minimum is \(k - n\).
\(n\) = the amount of random \(X\) variables in the set 1 to \(k\). Ex: If \(k = 100\) and there are 3 \(X\) variables (\(n = 3\)), then there are 97 possibilites for one of the variables to be the minimum since the other two would be larger.
Now to find how many ways the larger \(X\) variables can fit into the set 1 to \(k\). This would be \(k - (n - 1)\). Ex: if \(k = 10\) and there were 6 \(X\) variables (\(n = 6\)), there are 5 different possibilites where order matters:
So right now I’m at this point:
\(P(Y) = \frac{k-(n - 1)}{k-n}\)
Let’s say \(y\) is the minimum value. Now we only care about the permutations of the \(X\) variables that are larger than \(y\) (between \(y\) and \(k\)). So now the integer set is \(k-y\) times the number of permutations of \(n\) variables (\(k-(n-1)\)).
Final Solution:
\(P(Y) = \frac{(k-y)(k-(n - 1))}{k-n}\)
A. Geometric
Success = 0.1
Failure = 0.9
\(P(X \geq 9) = \sum_{n=9}^{\infty}(1-p)^{n-1}*p\)
\(P(8 \ failures \ in \ a \ row) = .61\)
\(E(X) = \frac{1}{p} = 10\)
\(V(X) = \frac{(1-p)}{p^2} = 90\)
\(\sigma = \sqrt{90} = 9.486\)
#Probability it will take 9 years to see the first failure
pgeom(8, prob = .1)## [1] 0.6125795B. Exponential
\(\mu = 10\) years
\(P(X \geq 9) = e^{\frac{-k}{\mu}} = e^{\frac{-9}{10}}\)
\(P(X \geq 9) = .406\)
\(E(X) = \frac{1}{\lambda} = \frac{1}{0.1} = 10\)
\(V(X) = \frac{1}{\lambda^2} = \frac{1}{0.1^2} = 100\)
\(\sigma = 10\)
m <- 10
k <- 9
prob <- exp(-k/m)
prob## [1] 0.4065697C. Binomial
Probability of 0 successes in 8 years
\[\left (\begin{array}{c} n\\ k \end{array} \right) = p^k(1-p)^{n-k}\]
\(p = .1\)
\(k = 0\)
\(n = 8\)
\[P(8) = \left (\begin{array}{c} 8\\ 0 \end{array} \right) = 0.1^8(1-0.1)^{8-0} = 0.1^0*0.9^8 = 0.43\]
#
#Probability of 0 successes in 8 years
dbinom(0, size = 8, prob = .1)## [1] 0.4304672D. Poisson
Time frame is 10 years. Probability of a failure is 1 in 10 years.
\(P(X) = \frac{\lambda^{x}e^{-\lambda}}{x!}\)
\(P(X \geq 9) = .37\)The cumulative probability between the first and eighth year subtracted from 1.
\(E(X) = \lambda = 1\) (average 1 failure out of 10 years)
\(\sigma = \sqrt{\lambda} = \sqrt{1} = 1\)
Find cumulative probability of a failure between the first and eighth years. Subtract it by one to find the probability of a failure after 8 years.
x <- 1
sum <- 0
while(x <= 8){
prob <- ((1^x)*exp(-1))/factorial(x)
sum <- sum + prob
x <- x + 1
print(prob)
}## [1] 0.3678794
## [1] 0.1839397
## [1] 0.06131324
## [1] 0.01532831
## [1] 0.003065662
## [1] 0.0005109437
## [1] 7.299195e-05
## [1] 9.123994e-061 - sum## [1] 0.3678806