(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? If the median cost (consider this the
best point estimate) is about 100,000 dollars per positive case total
and the test itself costs $1000 per administration, what is the total
first-year cost for treating 100,000 individuals?
Solution 1
1(a) Probability that an individual reported positive has the
disease. i.e. a true positive
Let event B = Positive test, A1 = Actual HIV pt, A2 = Actual non-HIV
P(A1) = 0.001; P(A2) = 0.999; P(B|A1) = 0.96
P(B|A2) = 1 - 0.98 = 0.02
Using Bayes formula
P(A1|B) = P(B|A1)P(A1) / P(B|A1)P(A1) + P(B|A2)P(A2)
p_true_positive = round((.96*.001)/((.96*.001)+(.02*.999)), 4)
p_true_positive## [1] 0.04581(b) Total first-year cost for treating 100,000 individuals
# $100,000 per positive case total and the test costs $1000 per administration
positive_cases_prevalence = 100000 * 0.001 # about 0.1% prevalence rate
cost_to_administer_test = 1000 * 100000 # it costs $1000 to administer each test
cost_to_treat_positive = positive_cases_prevalence * 100000 # cost to treat positive cases
total_cost = cost_to_administer_test + cost_to_treat_positive
total_cost## [1] 1.1e+08(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? What is the
probability that, after 24 months, you received 2 or more inspections?
What is the probability that your received fewer than 2 inspections?
What is the expected number of inspections you should have received?
What is the standard deviation?
Solution 2
Let p= probability of receiving a joint commission inspection in any
given month and q=1-p represent the probability of not receiving the a
joint inspection in any given month.
2(a) Probability of exactly 2 inspection after 24
months.
\(P(X = x) =
\binom{n}{x}p^{x}(1-p)^{n-x}\)
\(P(X=2) =
\binom{24}{2}(0.05^{2})(1-0.05)^{24-2} = 0.2232\)
Therefore, the probability of getting exactly 2 inspections after 24
months is 0.2232
# P(X=2) =
p = round(dbinom(2, 24, 0.05), 4)
p## [1] 0.2232 2(b) Probability of 2 or more inspections
P(X = 2 or more) = 1 - P(fewer than 2) = 1 - P(x = 0 or 1).
P(X = 2 or more) = 1 - P(X = 0) + P(X = 1)
P(X = 2 or more \(= 1-
[\binom{24}{0}(0.05^{0})(1-0.05)^{24-0} +
\binom{24}{1}(0.05^{1})(1-0.05)^{24-1}]=
1-(0.291989+0.368828)\)
= \(1-0.660817 = 0.339183 =
0.33918\)
Therefore, the probability of 2 or more inspections is about 0.3392
# P(X = 2 or more) =
p_2_or_more = round(1 - (dbinom(0, 24, 0.05) + dbinom(1, 24, 0.05)), 4)
p_2_or_more## [1] 0.3392 2(c) Probability of fewer than 2 inspections
P(X< 2) = P(X = 0) + P(X = 1) = \(\binom{24}{0}(0.05^{0})(1-0.05)^{24-0} +
\binom{24}{1}(0.05^{1})(1-0.05)^{24-1}\)
\(P(X < 2) = 0.291989 + 0.368828 = 0.660817
= 0.6608\)
Therefore the probability of fewer than 2 inspections is about
0.6608
# P(X < 2) =
p_less_than_2 = round(dbinom(0, 24, 0.05) + dbinom(1, 24, 0.05), 4)
p_less_than_2## [1] 0.6608(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? What is the probability that more
than 10 arrive in one hour? How many would you expect to arrive in 8
hours? What is the standard deviation of the appropriate probability
distribution? 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?
Solution 3
The poison distribution formula is given by:
\(f(x) = \frac{\lambda
^{x}}{x!}e^{-\lambda}\)
\(\lambda = 10/hour\)
3(a) Probability of exactly 3 in one hour
\(P(x=3) = \frac{10^{3}}{3!}e^{-10} =
0.0076\)
Therefore the probability of exactly 3 per hour is about 0.0076
# P(X=3) =
p = round(dpois(3, 10), 4)
p## [1] 0.0076
3(b) Probability that more than 10 arrive in an hour
P(X = more than 10) = 1 - P(X < 10) = 1 - P(X <= 9)
# The ppois function in R calculates the probability of a random variable that will be equal to or less than a number.
p_less_than_equal_9 = ppois(9, 10)
p_10_or_more = round((1 - p_less_than_equal_9), 4)
p_10_or_more## [1] 0.5421
3(b) How many would you expect to arrive in 8 hours
The expected value of a poison distribution is \(\lambda\). Therefore, we will expect 10 to
arrive in an hour and expect 10 x 8 = 80 to arrive in 8 hours.
3(c) Standard deviation
\(\sigma = \sqrt{\lambda}\)
\(=> \sqrt{10} = 3.1623\)
3(d) Percent Utilization if there are 3 family practice providers
who can see 24 patients daily
number_patients_seen_daily = 3 * 24
number_expected_to_arrive_8hours = 80
percent_utilization = (number_patients_seen_daily/number_expected_to_arrive_8hours) * 100
percent_utilization## [1] 90(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? How many nurses would we have
expected your subordinate to send? How many non-nurses would we have
expected your subordinate to send?
Solution 4
The hyper geometric distribution is given by:
\(P(X = x) =
\frac{\binom{m}{x}\binom{M-m}{k-x}}{\binom{M}{k}}\)
Where M = population size, m = number of success states in the
population,
k= quantity drawn in each trial and x = number of observed
successes.
M = Total number of workers supervised: 30
m = Number of nurses among those workers: 15
M- m = Number of non-nurses among those workers: 15
k = Total number of workers selected for the trips: 6
x = Number of nurses to be selected for the trips: 5
k - x = Number of non-nurses to be selected for the trips: 1
4(a) Probability the subordinate would have selected exactly
five nurses
\(P(X = 5) =
\frac{\binom{15}{5}\binom{30-15}{6-1}}{\binom{30}{6}} =
\frac{\binom{15}{5}\binom{15}{1}}{\binom{30}{6}}\)
\(=> P(X=5) = \frac{3003
*15}{593775}=0.0759\)
#P(5), x=5,m=15,n=15,k=6
p_5_nurses_1_non_nurse = round(dhyper(5,15,15,6,log=FALSE), 4)
p_5_nurses_1_non_nurse## [1] 0.0759 4(b) Expected number of nurses to send
Probability of sending nurses = 15/30 = 0.5; Number of trips = 6
Expected number of nurses to send = 6 * 0.5 = 3
Therefore, we would expect about 3 nurses to be sent on the trips.
4(c) Expected number of non-nurses to send
Probability of sending non-nurses = 15/30 = 0.5; Number of trips =
6
Expected number of non-nurses to send = 6 * 0.5 = 3
Therefore, we would expect about 3 non-nurses to be sent on the
trips.
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? In the course of 15 months? What is the expected number of
hours that a driver will drive before being seriously injured? Given
that a driver has driven 1200 hours, what is the probability that he or
she will be injured in the next 100 hours?
Solution 5
The geometric distribution is given by:
\(P(X=x) = p(1-p)^{x-1}\)
Where p = the probability of success and x = the number of trials up to
the first success. Basically this means the number of trials made upon
achieving the first success.
In this case, let us take the event of success as being seriously
injured in a car crash (Nobody sure wants this event as a success) but
we will use it in this case.
p = .1% = 0.001
5(a) Probability of getting seriously injured during the course of
the year (i.e 1200 hours of traverse)
x = 1200. This assumes that the driver will traverse 1200 hours before
being injured. \(P(X= 1200) =
(0.001)(1-0.001)^{1200-1} = 0.000301\)
p_injured_1200 = round(dgeom(1200, 0.001), 6)
p_injured_1200## [1] 0.000301 5(b) Probability of getting seriously injured during the
course of 15 months (i.e 1500 hours of traverse)
x = 1200. This assumes that the driver will traverse 1200 hours before
being injured. \(P(X= 1200) =
(0.001)(1-0.001)^{1500-1} = 0.000223\)
p_injured_1500 = round(dgeom(1500, 0.001), 6)
p_injured_1500## [1] 0.000223 5(c) What is the expected number of hours the driver will
drive before being seriously injured.
The expected values of a geometric distribution is the inverse of the
probability of success.
\(E(X) = 1/p = 1/0.001 = 1000
hours\)
5(d) Given that a driver has driven 1200 hours,what is the
probability that he or she will be injured in the next 100
hours
This means that we want to get the probability that the driver will be
injured after 1200 hours but before 1300 hours.
p(X=injured in the next 100 hours)= P(1300) - P(1200)
p_injured_next_100 = round((dgeom(1300, 0.001) - dgeom(1200, 0.001)), 6)
p_injured_next_100## [1] -2.9e-05You 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? What is the
expected value?
Solution 6
We will model this using the poison distribution.
rate = \(\lambda\) = 1 in 1000
6(a) Probability that the generator will fail more than twice in
1000 hours P(X = more than 2) = 1 - P(X <= 2)
p_less_than_equal_2 = ppois(2, 1)
p_more_than_2 = round((1 - p_less_than_equal_2), 4)
p_more_than_2## [1] 0.0803 6(b) Expected Value
The expected value is 1 in 1000.
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? 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? What is the expected waiting time?
Solution 7
This assumes a uniform distribution 7(a) Probability that the
patient will wait more than 10 minutes
# P(X > 10) = 1 - P(X<=10)
p_10mins_or_less = punif(10,0,30)
p_more_than_10mins = round((1 - p_10mins_or_less), 4)
p_more_than_10mins## [1] 0.6667
7(b) If the patient has already waited 10mins, what is the
probability that the patient will wait at least another 5
minutes
p_15mins = 1 - punif(15,10,30)
P_10mins = 1-punif(10,10,30)
p_atleast_another_5mins = round((p_15mins/P_10mins), 4)
p_atleast_another_5mins## [1] 0.757(c) Expected waiting time
expected_wait_time = (0 + 30)/2
expected_wait_time## [1] 15Your 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? What is the standard deviation? What is the
probability that your MRI will fail after 8 years? 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?
Solution 8
The exponential distribution is given by \(f(x) = \lambda e^{-\lambda x}\)
Expected value, E(x) = 10
8(a) What is the expected failure time
\(E(X) = \frac{1}{\lambda}\)
\(\lambda = 1/10 = 0.1/year\)
So, the parameter \(\lambda=0.1\);
Therefore the expected failure time is 10 years.
8(b) What is the Standard Deviation?
\(\sigma = \frac{1}{\lambda}\)
\(sigma = \frac{1}{0.1} = 10\)
Therefore, the standard deviation is 10 years as well.
8(c) Probability MRI will fail after 8 years
P(X > 8) = 1 - P(X<=8)
p_fail_after_8years = round((1 - pexp(8, 0.1)), 4)
p_fail_after_8years## [1] 0.44938(d) Probability that the MRI machine will fail in the next two years after owning it for 8 years.
p_fail_after_2years_given8 = round(((pexp(10,.1)-pexp(8,.1))*(1-pexp(8,.1)))/(1-pexp(8,.1)), 4)
p_fail_after_2years_given8## [1] 0.0814