Probability

Question 1:

# Prevalence Rate
Rate <- 0.001
# Sensitivity
P_Pos_Dis <- 0.96
# Specificity
P_Neg_notDis <- 0.98
P_Pos_notDis <- 1 - P_Neg_notDis

# Probability of testing positive
P_Pos <- P_Pos_Dis * Rate + P_Pos_notDis * (1 - Rate)

# Probability of having the disease given a positive test
P_D_Pos <- P_Pos_Dis * Rate / P_Pos

# Median cost per positive case
cost_per_positive <- 100000
# Cost per test administration
test_cost <- 1000
# Number of individuals
num_individuals <- 100000

# Total cost for treating 100,000 individuals
total_cost <- (P_D_Pos * cost_per_positive * num_individuals) + (num_individuals * test_cost)

P_D_Pos

## [1] 0.04584527

total_cost

## [1] 558452722

Question 2

# Probability of receiving a Joint Commission inspection in any given month
p <- 0.05
# Number of months
n <- 24

# Probability of receiving exactly 2 inspections
P2 <- dbinom(2, size = n, prob = p)

# Probability of receiving 2 or more inspections
P2more <- 1 - pbinom(1, size = n, prob = p)

# Probability of receiving fewer than 2 inspections
P2less <- pbinom(1, size = n, prob = p)

# Expected number of inspections
expected <- n * p

# Standard deviation
sd <- sqrt(n * p * (1 - p))

P2

## [1] 0.2232381

P2more

## [1] 0.3391827

P2less

## [1] 0.6608173

expected

## [1] 1.2

sd

## [1] 1.067708

Question 3

# Probability that exactly 3 arrive in one hour
P3 <- dpois(3, 10)

# Probability that more than 10 arrive in one hour
P10 <- 1 - ppois(10, 10)

# Expected number of arrivals in 8 hours
expected <- 10 * 8

# Standard deviation
sd <- sqrt(10 * 8)

# Percent utilization
utilization <- (expected / (24 * 3)) * 100

P3

## [1] 0.007566655

P10

## [1] 0.4169602

expected

## [1] 80

sd

## [1] 8.944272

utilization

## [1] 111.1111

The percentage of utilization is greater than 100%. I’d recommend employing more provider.

Question 4

# Probability of selecting 5 nurses innocently
P <- dhyper(5, m = 15, n = 15, k = 6)

# Expected number of nurses selected
expected_nurses <- 6 * (15 / 30)

# Expected number of non-nurses selected
expected_non_nurses <- 6 * (15 / 30)

P

## [1] 0.07586207

expected_nurses

## [1] 3

expected_non_nurses

## [1] 3

Question 5

# Probability of being seriously injured in a car crash during the course of a year
Pyear <- 1- pgeom(1200,.001)

# Probability of being seriously injured in a car crash during the course of 15 months
P15months <- 1-pgeom(1500,.001)

# Expected number of hours that a driver will drive before being seriously injured
expected <- 1 / 0.001

# Probability of being injured in the next 100 hours, given that the driver has already driven for 1200 hours

P100hours <- 1- pgeom(100,.001)

Pyear

## [1] 0.3007124

P15months

## [1] 0.2227398

expected

## [1] 1000

P100hours

## [1] 0.9038874

Question 6

# Failure rate of the generator
lambda <- 1/1000  

# Probability that the generator will fail more than twice in 1000 hours
P <- 1 - ppois(2, lambda * 1000)

# Expected value of generator failures in 1000 hours
expected_failures <- lambda * 1000

P

## [1] 0.0803014

expected_failures

## [1] 1

Question 7

# Probability that the patient will wait more than 10 minutes
P10min <- (30 - 10) / (30 - 0)

# Probability that the patient will wait at least another 5 minutes after waiting 10 minutes
P5min <- ((30 - 15) / (30 - 0))/((30 - 10) / (30 - 0))

# Expected waiting time 
expected <- (30 - 0) / 2

P10min

## [1] 0.6666667

P5min

## [1] 0.75

expected

## [1] 15

Question 8

# Expected failure time 
expected <- 10

# Standard deviation
sd <- 10

# Probability that the MRI will fail after 8 years
P8 <- pexp(8, rate = 1/10, lower.tail = FALSE)

# Probability that the MRI will fail in the next two years, given that it has been owned for 8 years
P2 <- pexp(10, rate = 1/10) - pexp(8, rate = 1/10)

expected

## [1] 10

sd

## [1] 10

P8

## [1] 0.449329

P2

## [1] 0.08144952

…