\(P(A_k|B) = \frac {P(B|A_k)*P(A_k)}{\sum_{i = 1}^{k}{P(B|A_i)*P(A_i)}}\)
In this case, \(A_k\) will represent correct test results. \(P(B|A_{pos}) = 0.96\) \(P(A_{pos}) = 0.001\) \(P(B|A_{neg}) = 0.02\) \(P(A_{neg}) = 0.999\)
prob <- ((.96)*(.001))/((.96)*(.001) + (.02)*(.999))
round(prob,3)## [1] 0.046cost <- 100000*(0.001*100000) + 1000*100000
round(cost,3)## [1] 1.1e+08Ignoring test results and considering the actual prevalence of the disease, the expected cost would be $110,000,000 for the first year of treatment and testing for 100,000 individuals.
The probability of receiving exactly 2 inspections can be represented by \(P(exactly two) = \binom{24}{2} * p^2*(1 - p)^{22}\)
p <- .05
round((choose(24,2)*(p^2)*((1-p)^22)), 6)## [1] 0.223238P(exactly 2) = 0.223238
\(P(\geq 2) = 1 - P(one) - P(none)\)
p <- .05
p1 <- (choose(24,1)*(p^1)*((1-p)^23))
p0 <- (choose(24,0)*(p^0)*((1-p)^24))
round(1 - (p1 + p0), 6)## [1] 0.339183P(Two or more) = 0.339183
\(P(\leq 2) = P(one) + P(none)\)
This is also the complement of having two or more inspections.
p <- .05
p1 <- (choose(24,1)*(p^1)*((1-p)^23))
p0 <- (choose(24,0)*(p^0)*((1-p)^24))
round((p1 + p0), 6)## [1] 0.660817P(Fewer than two) = 0.660817
\(E = 24*p\)
round(24*p,4)## [1] 1.2The expected number of inspections is 1.2, or 1.
\(\sigma = \sqrt{np(1 - p)}\)
round((24*.05*(.95))^(1/2), 3)## [1] 1.068Standard Deviation = 1.068
The PMF for a Poisson Distribution is \[\frac {e^{-\lambda} \lambda^x}{x!}\].
lambda <- 10
x <- 3
pmf <- (exp(-lambda)*lambda^x)/(factorial(x))
round(pmf,5)## [1] 0.00757Probability of Exactly 3: 0.00757
The probability of more than 10 coming can be represented by \(1 - \sum P(0):P(10)\)
prob <- 1
lambda <- 10
pmf <- function(lambda,x){
out <- (exp(-lambda)*(lambda^x))/(factorial(x))
return(out)
}
for(i in 0:10){
prob <- prob - pmf(10, i)
}
prob## [1] 0.4169602Probability of More than 10: 0.41696
The expectation in one hour is \(\lambda = 10\), so the expectation for 8 hours is simply \(\lambda \times 8 = 80\)
The standard deviation is the square root of the variance. For a poisson distribution, the variance is \(\lambda\). So, Standard Deviation = \(\sqrt {10} = 3.16228\)
The percent utilization is given by dividing the expectation by the utilization capacity.
use <- 80/(24*3)
use## [1] 1.111111These doctors are at 111% capacity. They should add another doctor or add a part time doctor to cover the extra patients in order to appropriately meet the demands of the practice.
At a glance, this does appear to be favoritism.
The PMF for a hypergeometric distribution is \(\frac {\left[ {\binom{m}{x} {\binom{N - m}{n - x}}} \right]} {\binom{N}{n}}\). In this case, m will represent the number of nurses and n will represent the number of trips. x represents the target quantity of 5 nurses.
phyper <- (choose(15, 5)*choose(30 - 15, 6 - 5))/choose(30,6)
phyper## [1] 0.07586207Probability that it occurred by chance: 0.07586
The mean, or expectation for a hypergeometric distribution is given by \(n \times \frac m N\), where \(n\) is the number of trips, m is the number of nurses, and N is the total population.
Thus, the expected amount of nurses is \(n \times \frac m N = 6 \times \frac {15}{30} = 3\) The same can be done to compute the number of non-nurses that are expected. Expectation of non-nurses: \(n \times \frac m N = 6 \times \frac {15}{30} = 3\)
The PMF for a geometric distribution is \((1-p)^{x - 1}p\) where x is the number of trials and p is the probability of the event happening in one trial. The probability of at least one serious injury in the first year can be found by subtracting the probability that there is no serious injury from 1.
trials <- 1200
p <- .001
pgeom <- 1 - ((1 - p)^(trials))
pgeom## [1] 0.6989866It is very likely that a serious injury occurs in a year. Probability of injury within a year: 0.69899
trials <- 1200*(15/12)
p <- .001
pgeom <- 1 - ((1 - p)^(trials))
pgeom## [1] 0.7770372expect <- 1/p - 1
expect## [1] 999Probability of injury within 15 months: 0.77704
The expected number of hours to drive before injury can be found from \(E[X] = \frac 1p - 1\) Expectation: 999 hours before serious injury (Injury would occur in the 1000th hour)
Probability of injury occuring between 1300 and 1200 hours:
The important idea here is the memoryless property that applies to geometric distributions. Memorylessness means that regardless of the history of a variable, the future remains independent. We are given that a driver has driven 1200 hours. The probability that they will be injured in the next 100 hours is simply the probability of the injury occurring in 100 hours, regardless of how many hours they have already driven.
trials <- 100
p <- .001
pgeom <- 1 - ((1 - p)^(trials))
pgeom## [1] 0.09520785Thus, the probability of an injury occurring in the next 100 hours is 0.09521.
This appears to be a Poisson distribution.
prob <- 1
lambda <- 1
pmf <- function(lambda,x){
out <- (exp(-lambda)*(lambda^x))/(factorial(x))
return(out)
}
for(i in 0:2){
prob <- prob - pmf(lambda, i)
}
prob## [1] 0.0803014Probability of more than two failures: 0.0803
Because this is likely a Poisson random variable, the expectation is equivalent to \(\lambda\) = 1.
Under a continuous uniform distribution, the PDF is \(\frac 1{b-a}\), where \(a\) and \(b\) are the boundaries, in this case 0 and 30. The probability that the patient waits more than 10 minutes is found via simple integration. The area under a uniform distribution forms a basic rectangle and the area of the rectangle is the probability. For a probability of more than ten minutes, we are looking for the area from 10:30.
len <- 30 - 10
hgt <- 1/30
p <- len*hgt
p## [1] 0.6666667Thus, the probability of waiting more than 10 minutes is 0.667 or \(\frac23\)
Probability of waiting to at least 15 minutes given already waiting 10 minutes:
\(P(A|B) = \frac{P(A\cap B)}{P(B)}\)
In this case, we do not need this equation. Given that the patient waited 10 minutes already, the probabilities of future events have changed. There are now only 20 more minutes in the interval, so the distribution needs to be adjusted.
The rectangle here shrinks because we are given that the patient already waited 10 minutes. The individual rectangles have now shrunk to \(\frac{1}{20}\) each.
hgt <- 1/20
p15 <- (hgt)*(20 - 5)
p15## [1] 0.75Probability of waiting at least 5 more minutes: 0.75
The expectation of a uniform distribution is \(\frac{a + b}{2}\), where a and b are the boundaries of the interval.
Expectation: 15 minutes
The expectation of an exponential distribution is \(\theta\), where \(\theta = \frac 1{\lambda}\). In this case, there is no rate parameter but the expectation is given to be 10 years. Expected failure time: 10 years
The standard deviation of an exponential distribution is \(\sqrt{\theta^2} = \theta\) Standard Deviation = 10
We need to evaluate the PDF at x = 8 years.
theta <- 10
x <- 8
pdfexp <- (exp(-(x/theta)))
pdfexp## [1] 0.449329Probability of failure at 8 years: 0.449329
Here, we need to evaluate the CDF from 8 to 10 years. CDF: \(1 - e^{- \frac {x}{\theta}}\).
This is not a traditional conditional probability question because the exponential distribution has the memoryless property. Thus, we need to find the probability of failure between 8 and 10 years.
cdfexp <- function(theta, x){
out <- 1 - exp(-x/theta)
return(out)
}
theta <- 10
up <- cdfexp(theta,10)
low <- cdfexp(theta,8)
pexp <- up - low
pexp## [1] 0.08144952Probability of failure between 8 and 10 years: 0.0814495