12 Named Discrete Distributions
12.1 Objectives
Recognize and set up for use common discrete distributions (Uniform, Binomial, Poisson, Hypergeometric) to include parameters, assumptions, and moments.
Use
R
to calculate probabilities and quantiles involving random variables with common discrete distributions.
12.2 Homework
For each of the problems below, 1) define a random variable that will help you answer the question, 2) state the distribution and parameters of that random variable; 3) determine the expected value and variance of that random variable, and 4) use that random variable to answer the question.
We will demonstrate using 1a and 1b.
12.2.1 Problem 1
The T-6 training aircraft is used during UPT. Suppose that on each training sortie, aircraft return with a maintenance-related failure at a rate of 1 per 100 sorties.
- Find the probability of no maintenance failures in 15 sorties.
\(X\): the number of maintenance failures in 15 sorties.
\(X\sim \textsf{Bin}(n=15,p=0.01)\)
\(\mbox{E}(X)=15*0.01=0.15\) and \(\mbox{Var}(X)=15*0.01*0.99=0.1485\).
\(\mbox{P}(\mbox{No maintenance failures})=\mbox{P}(X=0)={15\choose 0}0.01^0(1-0.01)^{15}=0.99^{15}\)
0.99^15
## [1] 0.8600584
## or
dbinom(0,15,0.01)
## [1] 0.8600584
This probability makes sense, since the expected value is fairly low. Because, on average, only 0.15 failures would occur every 15 trials, 0 failures would be a very common result. Graphically, the pmf looks like this:
gf_dist("binom",size=15,prob=0.01) %>%
gf_theme(theme_classic())
- Find the probability of at least two maintenance failures in 15 sorties.
We can use the same \(X\) as above. Now, we are looking for \(\mbox{P}(X\geq 2)\). This is equivalent to finding \(1-\mbox{P}(X\leq 1)\):
## Directly
1-(0.99^15 + 15*0.01*0.99^14)
## [1] 0.009629773
## [1] 0.009629773
## [1] 0.009629773
## or
1-pbinom(1,15,0.01)
## [1] 0.009629773
## or
pbinom(1,15,0.01,lower.tail = F)
## [1] 0.009629773
- Find the probability of at least 30 successful (no mx failures) sorties before the first failure.
\(X\): the number of maintenance failures out of 30 sorties.
\(X\sim \textsf{Binom}(n=30,p=0.01)\), and \(\mbox{E}(X)=0.3\) and \(\mbox{Var}(X)=0.297\).
\(\mbox{P}(\mbox{0 failures})=\mbox{P}(X=0)=0.99^{30}\)
0.99^30
## [1] 0.7397004
##or
dbinom(0,30,0.01)
## [1] 0.7397004
Using negative binomial, which was not in the reading but you can research:
\(Y\): the number of successful sorties before the first failure.
\(Y\sim \textsf{NegBin}(n=1,p=0.01)\), and \(\mbox{E}(X)=99\) and \(\mbox{Var}(X)=9900\).
\(\mbox{P}(\mbox{at least 30 successes before first failure})=\mbox{P}(Y\geq 30)\)
1-pnbinom(29,1,0.01)
## [1] 0.7397004
- Find the probability of at least 50 successful sorties before the third failure.
Using a binomial random variable, we have 52 trials and need at least 50 to be a success. The random variable is \(X\) the number of successful sorties out of 52.
1-pbinom(49,52,.99)
## [1] 0.9846474
Or using a negative binomial, let
\(Y\): the number of successful sorties before the third failure.
\(Y\sim \textsf{NegBin}(n=3,p=0.01)\), and \(\mbox{E}(X)=297\) and \(\mbox{Var}(X)=29700\).
\(\mbox{P}(\mbox{at least 50 successes before 3rd failure})=\mbox{P}(Y\geq 50)\)
1-pnbinom(49,3,0.01)
## [1] 0.9846474
Notice if the question had been exactly 50 successful sorties before the 3 failure, that is a different question. Then we could use either:
dbinom(50,52,.99)*.01
## [1] 0.000802238
The \(0.01\) is because the last trial is a failure.
Or
dnbinom(50,3,0.01)
## [1] 0.000802238
12.2.2 Problem 2
On a given Saturday, suppose vehicles arrive at the USAFA North Gate according to a Poisson process at a rate of 40 arrivals per hour.
- Find the probability no vehicles arrive in 10 minutes.
\(X\): number of vehicles that arrive in 10 minutes
\(X\sim \textsf{Pois}(\lambda=40/6=6.67)\) and \(\mbox{E}(X)=\mbox{Var}(X)=6.67\).
\(\mbox{P}(\mbox{no arrivals in 10 minutes})=\mbox{P}(X=0)=\frac{6.67^0 e^{-6.67}}{0!}=e^{-6.67}\)
exp(-40/6)
## [1] 0.001272634
##or
dpois(0,40/6)
## [1] 0.001272634
- Find the probability at least 50 vehicles arrive in an hour.
\(X\): number of vehicles that arrive in an hour
\(X\sim \textsf{Pois}(\lambda=40)\) and \(\mbox{E}(X)=\mbox{Var}(X)=40\).
\(\mbox{P}(\mbox{at least 50 arrivals in 1 hour})=\mbox{P}(X\geq 50)\)
1-ppois(49,40)
## [1] 0.07033507
- Find the probability that at least 5 minutes will pass before the next arrival.
\(X\): number of vehicles that arrive in 5 minutes
\(X\sim \textsf{Pois}(\lambda=40/12=3.33)\) and \(\mbox{E}(X)=\mbox{Var}(X)=3.33\).
\(\mbox{P}(\mbox{no arrivals in 5 minutes})=\mbox{P}(X=0)=\frac{3.33^0 e^{-3.33}}{0!}=e^{-3.33}\)
exp(-40/12)
## [1] 0.03567399
##or
dpois(0,40/12)
## [1] 0.03567399
12.2.3 Problem 3
Suppose there are 12 male and 7 female cadets in a classroom. I select 5 completely at random (without replacement).
- Find the probability I select no female cadets.
\(X\): number of female cadets selected out of sample of size 5
\(X\sim \textsf{Hypergeom}(m=7,n=12,k=5)\) and \(\mbox{E}(X)=1.842\) and \(\mbox{Var}(X)=0.905\).
\[ \mbox{P}(\mbox{no female cadets selected})=\mbox{P}(X=0)=\frac{{7\choose 0}{12\choose 5}}{{19\choose 5}} \]
## [1] 0.06811146
##or
dhyper(0,7,12,5)
## [1] 0.06811146
- Find the probability I select more than 2 female cadets.
Using the same random variable: \[ \mbox{P}(\mbox{more than 2 female})=\mbox{P}(X>2)=1-\mbox{P}(X\leq 2) \]
1-phyper(2,7,12,5)
## [1] 0.2365841
## [1] 0.2365841