1.

(Bayesian). A new test for multinucleoside-resistant (MNR) human immunodeficiency virus type 1 (HIV-1) variants was recently developed. The test maintains 96% sensitivity, meaning that, for those with the disease, it will correctly report “positive” for 96% of them. The test is also 98% specific, meaning that, for those without the disease, 98% will be correctly reported as “negative.” MNR HIV-1 is considered to be rare (albeit emerging), with about a .1% or .001 prevalence rate.

Given the prevalence rate, sensitivity, and specificity estimates, what is the probability that an individual who is reported as positive by the new test actually has the disease?

Answer: 0.04584527

# If we have 100000 people, the matrix below applies.
matrix1 <- matrix(c('', 'Has HIV', 'No HIV', 'Total', 'Positive', '96', '1998', '2094', 'Negative', '4', '97902', '97906', '', '100', '99900', '100000'), nrow = 4, ncol = 4, byrow = TRUE)
matrix1
##      [,1]       [,2]      [,3]     [,4]    
## [1,] ""         "Has HIV" "No HIV" "Total" 
## [2,] "Positive" "96"      "1998"   "2094"  
## [3,] "Negative" "4"       "97902"  "97906" 
## [4,] ""         "100"     "99900"  "100000"
# The number of people who are positive:
matrix1[2,4]
## [1] "2094"
# The number of positive people who have the disease:
matrix1[2,2]
## [1] "96"
# Probability:
as.integer(matrix1[2,2]) / as.integer(matrix1[2,4])
## [1] 0.04584527
# P(A | B) = P(B|A)P(A) / SUM(P(B|A)P(A))
# A = Has HIV
# B = Positive

# Bayesian Formula:
(0.96 * 0.001) / ((0.96 * 0.001) + (0.02 * 0.999))
## [1] 0.04584527

If the median cost (consider this the best point estimate) is about $100,000 per positive case total and the test itself costs $1000 per administration, what is the total first-year cost for treating 100,000 individuals?

Answer: The total cost will be $309,400,000 (if the prevalence, sensitivity, and specificity apply to this scenario).

Other: If all 100,000 people are positive, then the cost will be $10,100,000,000.

# If we assume that out of 100,000 individuals, only 2094 of them will be positive, then we have to account for the cost of the tests and the treatment for each person.
totalPeople <- 100000
costPerTest <- 1000
positivePeople <- 2094
costPerTreatment <- 100000
(totalPeople * costPerTest) + (positivePeople * costPerTreatment)
## [1] 309400000
# However, if we assume that all 100,000 people will need to have treatment, then we will multiply the number of people by the cost of both the test and treatment.
totalPeople * (costPerTest + costPerTreatment)
## [1] 1.01e+10

2. (Binomial).

The probability of your organization receiving a Joint Commission inspection in any given month is .05.

What is the probability that, after 24 months, you received exactly 2 inspections?

Answer: 0.2232381

library(pracma)
## Warning: package 'pracma' was built under R version 4.3.2
nchoosek(24, 2) * (0.05 ** 2) * (0.95 ** (24 - 2))
## [1] 0.2232381

What is the probability that, after 24 months, you received 2 or more inspections?

Answer: 0.3391827

1 - pbinom(1, 24, 0.05)
## [1] 0.3391827

What is the probability that you received fewer than 2 inspections?

Answer: 0.6608173

pbinom(1, 24, 0.05)
## [1] 0.6608173

What is the expected number of inspections you should have received?

Answer: 1.2 inspections (in real life that would be 1)

24 * 0.05
## [1] 1.2

What is the standard deviation?

Answer: 1.067708

sqrt(24 * 0.05 * (1 - 0.05))
## [1] 1.067708

3. (Poisson).

You are modeling the family practice clinic and notice that patients arrive at a rate of 10 per hour.

What is the probability that exactly 3 arrive in one hour?

Answer: 0.007566655

dpois(3, 10)
## [1] 0.007566655

What is the probability that more than 10 arrive in one hour?

Answer: 0.4169602

1 - sum(dpois(0:10, 10))
## [1] 0.4169602

How many would you expect to arrive in 8 hours?

Answer: 80 people

10 * 8
## [1] 80

What is the standard deviation of the appropriate probability distribution?

Answer: 3.162278 (without a time frame)

Other: 8.944272 (for 8 hours)

sqrt(10)
## [1] 3.162278
sqrt(10 * 8)
## [1] 8.944272

If there are three family practice providers that can see 24 templated patients each day, what is the percent utilization and what are your recommendations?

Answer: 111% utilization. . . Since this is more than 100%, there are not enough providers for the expected number of patients. You could consider hiring another doctor, or you could let some patients know that you need to refer them to a new practice. Make sure to also not accept any new patients. Another option is to increase the number of hours in the office (9 hours instead of 8 where no extra patients can come the last hour).

# In 8 hours, we expect 80 people. If we have 3 providers who can see 24 patients per day, that means... they will see 72 patients.
3 * 24
## [1] 72
# Utilization is 80 / 72
80 / 72
## [1] 1.111111
# 111%

4. (Hypergeometric).

Your subordinate with 30 supervisors was recently accused of favoring nurses. 15 of the subordinate’s workers are nurses and 15 are other than nurses. As evidence of malfeasance, the accuser stated that there were 6 company-paid trips to Disney World for which everyone was eligible. The supervisor sent 5 nurses and 1 non-nurse.

If your subordinate acted innocently, what was the probability he/she would have selected five nurses for the trips?

Answer: 0.07586207

#dhyper(k, K, N - K, n)
dhyper(5, 15, 15, 6)
## [1] 0.07586207

How many nurses would we have expected your subordinate to send?

Answer: 3 nurses

6 * (15/30)
## [1] 3

How many non-nurses would we have expected your subordinate to send?

Answer: 3 nurses

6 * 15/30
## [1] 3

5. (Geometric).

The probability of being seriously injured in a car crash in an unspecified location is about .1% per hour. A driver is required to traverse this area for 1200 hours in the course of a year.

What is the probability that the driver will be seriously injured during the course of the year?

Answer: 0.69899866

(1 - (1 - 0.001)^1200)
## [1] 0.6989866

In the course of 15 months?

Answer: 0.7770372

fifteenMonths <- 1200*1.25
(1 - (1 - 0.001)^fifteenMonths)
## [1] 0.7770372

What is the expected number of hours that a driver will drive before being seriously injured?

Answer: 1000 hours

1 / 0.001
## [1] 1000

Given that a driver has driven 1200 hours, what is the probability that he or she will be injured in the next 100 hours?

Answer: 0.09520785

(1 - (1 - 0.001)^100)
## [1] 0.09520785

6.

You are working in a hospital that is running off of a primary generator which fails about once in 1000 hours.

What is the probability that the generator will fail more than twice in 1000 hours?

Answer: 0.997

1 - sum((1 - (1/1000))^(0:2 - 1) * (1/1000))
## [1] 0.997

What is the expected value?

Answer: 1000 hours

1 / (1/1000)
## [1] 1000

7.

A surgical patient arrives for surgery precisely at a given time. Based on previous analysis (or a lack of knowledge assumption), you know that the waiting time is uniformly distributed from 0 to 30 minutes.

What is the probability that this patient will wait more than 10 minutes?

Answer: 2/3

1 - punif(10, 0, 30)
## [1] 0.6666667

If the patient has already waited 10 minutes, what is the probability that he/she will wait at least another 5 minutes prior to being seen?

Answer: 0.5

1 - punif(15, 0, 30)
## [1] 0.5

What is the expected waiting time?

Answer: 15 minutes

(0 + 30) / 2
## [1] 15

8.

Your hospital owns an old MRI, which has a manufacturer’s lifetime of about 10 years (expected value). Based on previous studies, we know that the failure of most MRIs obeys an exponential distribution.

What is the expected failure time?

Answer: 10 years

1 / (1/10)
## [1] 10

What is the standard deviation?

Answer:

1 / (1/10)
## [1] 10

What is the probability that your MRI will fail after 8 years?

Answer: 0.550671

1 - exp((-1/10) * 8)
## [1] 0.550671

Now assume that you have owned the machine for 8 years. Given that you already owned the machine 8 years, what is the probability that it will fail in the next two years?

0.1812692

1 - exp((-1/10) * 2)
## [1] 0.1812692