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.
Step1: The probability of \(X_i\) to be greater than a value, let’s say \(l\), where \(1 \leq l < k\) will be the number of outcomes greater than \(l\) divided by the total number of outcomes which is \(k\). Since there are \(n\) independent random variables, then:
\(P(X_i > l) = (\frac {k-l}{k})^n\) (1)
Now, to find the probability that the minimum of all \(X_i\)s (denoted by Y) is equal to \(l\) is:
\(P(Y=l) = P(Y>l-1) - P(Y>l)\) (2)
Using the equation (1), we can substitute into equation (2):
\(P(Y=l) = \left(\frac{k-(l-1)}{k} \right)^n - \left(\frac{k-l}{k} \right)^n\)
\(\Rightarrow P(Y=1) = \left(\frac{k-l+1}{k} \right)^n - \left(\frac{k-l}{k} \right)^n\)
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.).
\(\Rightarrow\) The solution:
The probability that the MRI will fail in any year during the 10 years is \(p=\frac{1}{10}\), then the probability of not failing will be \(q=1-p = 1-\frac{1}{10}=\frac{9}{10}\)
So, \(P(\text{not failing during the 8 years}) = q^8 \times p = (\frac{9}{10})^8 \times \frac{1}{10}\)
# Define the probability of failure in a given year
p <- 1 / 10
# Probability of not failing
q <- 9 / 10
# Probability of not failing in 8 years
P_fail_in_8 <- (q^8) * p
cat("Probability of not failing during the 8 years is :", P_fail_in_8, "\n")## Probability of not failing during the 8 years is : 0.04304672The expected value (the mean) will be just 10 years, since that is the lifetime of the MRI machine.
The standard deviation is \(\sigma = \frac {\sqrt{1-p}}{p}\)
SD_value <- sqrt((1 - p) / p^2)
cat("Standard deviation:", SD_value, "years\n")## Standard deviation: 9.486833 years\(\Rightarrow\) The solution:
The probability that the MRI will fail after 8 years (before its lifetime) is \(1-P(\text{not failing during the 8 years})\)
\(P(\text{not failing during the 8 years}) = e^{-\lambda \cdot 8}\) where \(\lambda = 1/10\)
eight_year_fail <- 1- exp((-1/10)*8)
cat("The probability that the MRI will fail after 8 years is", eight_year_fail)## The probability that the MRI will fail after 8 years is 0.550671The expected failure time is the expected value; 10 years.
Since this is an exponential distribution, the standard deviation is the same as the mean; so \(SD=10\)
\(\Rightarrow\) The solution:
\(\Rightarrow\) The binomial formula is: \(P(X=k)=\binom{n}{k} p^k (1-p)^{n-k}\) Where: \(n=8\), $ k=0$ and given the rate of the failure is only once per 10 years, then the \(P(\text{failure}) =0.1\) while \(P(\text{success}) = 0.9\)
Then; $P() = p^0 (1-p)^{8} = (0.9)^0 (0.1)^{8} $
And: the expected value is: \(E=n \times p = 8 \times 0.9\)
The standard deviation is: \(\sigma =\sqrt {n \times p \times (1-p)}\)
# Defining the coefficients:
n <- 8
k <- 0
p <- 0.9
# The probability of failing after 8 years
Fail_after_8years <- (p^0 ) * ((1-p)^8)
# Expected Value
E_Value <- n * p
# Standard deviation
StandardV <- (n*p*(1-p))^(1/2)
# Display all values calculated above
cat ("The probability that the machine will fail after 8 years is ", Fail_after_8years, "\n", "The Expected Value is ", E_Value, "years", "\n", "The standard Deviation is ", StandardV, "years", "\n")## The probability that the machine will fail after 8 years is 1e-08
## The Expected Value is 7.2 years
## The standard Deviation is 0.8485281 years\(\Rightarrow\) The solution:
The Poisson formula is \(P(X=k) = \frac{\lambda^k}{k!} e^{-\lambda}\)
So, the probability that the machine will fail after 8 years is: \(P(\text{failing after 8 years} = 1 - P(\text{not failing in 8 years}))\) And \(P(\text{not failing in 8 years} = \sum_{k=0}^{8} \frac {\lambda^k}{k!} e^{-\lambda}\)
The Expected value is \(E(X) = \lambda\) and the standard deviation is \(\sigma = \sqrt {\lambda}\)
# Define average
lambda <- 8
# Calculate the probability of not failing
for (i in 0:8) {
prob <- (lambda^i / factorial(i)) * exp(-lambda)
}
# Probability of failing
prob_fail <- 1-prob
cat("Probability of machine failing after 8 years:", prob_fail, "\n")## Probability of machine failing after 8 years: 0.8604135# Expected value and standard deviation
ExpectedV <- lambda
SD_Value1 <- sqrt(lambda)
cat ("The expected value is ", ExpectedV, "\n" , "The standard Deiviation is ", SD_Value1, "\n")## The expected value is 8
## The standard Deiviation is 2.828427