A r.v. X is said have a Bernoulli distribution with success parameter p if P(X=0)=1-p and P(X=1)=p.
Note: often you see q = 1-p used
Shorthand: \(X \sim Ber(p)\)
we already saw that then EX = p and Var(X) = pq
Say X1, … , Xn are iid Ber(p) then
\[ Y = \sum_{i-1}^n X_i \] is said to have a Binomial distribution with parameters n and p. We have
\[ Y \sim Bin(n, p) \\ P(Y=k) = \begin{pmatrix} n \\ k \end{pmatrix}p^k(1-p)^{n-k}, k=0,..,n \\ E[Y] = np \\ Var[Y] = np(1-p) \]
A company wants to hire 5 new employees. From previous experience they know that about 1 in 10 applicants are suitable for the jobs. What is the probability that if they interview 20 applicants they will be able to fill those 5 positions?
Consider each interview a “trial” with the only two possible outcomes: “success” (can be hired) or “failure” (not suitable).
Assumptions:
“success probability” is the same for all applicants (as long as we know nothing else about them this is ok.)
trials are independent (depends somewhat on the setup of the interviews but should be ok)
then if we let Y = “#number of suitable applicants in the group of 20” we have \(Y \sim B(20, 0.1)\) and
\[ P(Y \ge 5)=1-P(Y\le4) \]
1-pbinom(4, 20, 0.1)## [1] 0.04317Say X1, X2, .. are iid Ber(p) and let Y be the number of trials needed until the first success. Then Y is said to have a geometric distribution with rate p, \(Y\sim G(p)\), and we have
\[ Y \sim G(p) \\ P(Y=k) = p(1-p)^{k-1}, k=1,2,.. \\ E[Y] = 1/p \\ Var[Y] = (1-p)/p^2 \]
(same as above) How many applicants will the company need to interview to be 90% sure to be able to fill at least one of the five positions?
if we let Y be the number of trials until the first success (= an applicant is suitable) we have \(Y\sim G(0.1)\). Then
\[ \begin{aligned} &0.9=P(Y \le n)= \\ &\sum_1^n p(1-p)^{k-1}=\\ &p\sum_0^{n-1} (1-p)^k=\\ &p \frac{1-(1-p)^n}{1-(1-p)}=\\ &1-(1-p)^n \\ &(1-p)^n=0.1\\ &n\log (1-p)=\log (0.1)\\ &n=\frac{\log (0.1)}{\log (1-p)}=\\ &\frac{\log (0.1)}{\log (1-0.1)}=21.8 \end{aligned} \]
qgeom(0.9, 0.1) + 1## [1] 22Note The command geom in R is for a r.v. \(Y^* = Y-1\) (the number of failures until the first success), that is it takes values 0,1, … instead of 1,2,.. and \(P(Y^*=k)=P(Y=k+1)\).
Despite the different name this is actually a generalization of the geometric, namely where Y is the number of trials needed until the rth success.
\[ Y \sim NB(r,p) \\ P(Y=k) = \begin{pmatrix} k-1 \\ r-1 \end{pmatrix}p^r(1-p)^{k-r}, k=r, r+1,..\\ E[Y] = r/p \\ Var[Y] = r(1-p)/p^2 \]
Note as with the geometric the R function nbinom uses a slightly different parametrization, it is for a r.v. \(Y^* = Y-r\).
(same as above) How many applicants will the company need to interview to be 90% sure to be able to fill all of the five positions?
if we let Y be the number of trials until the 5th success we have \(Y\sim NB(0.1, 5)\). Then
qnbinom(0.9, 5, 0.1) + 5## [1] 78(Note: it is not \(5*20=100\)!)
Consider an urn containing N+M balls, of which N are white and M are black. If a sample of size n is chosen at random and if Y is the number of white balls chosen, then Y has a hypergeometric distribution with paramaters (n,N,M).
\[ Y \sim HG(n,N,M) \\ P(Y=k) = \begin{pmatrix} N \\ k \end{pmatrix} \begin{pmatrix} M \\ n-k \end{pmatrix}/ \begin{pmatrix} N+M \\ n \end{pmatrix}, k=0,..,n \\ E[Y] = \frac{nN}{N+M} \\ Var[Y] = \frac{nN}{(N+M)^2}\left( 1-\frac{n-1}{N+M-1} \right) \]
say our company has a pool of 100 candidates for the job, 10 of whom are suitable for hiring. If they interview 50 of the 100, what is the probability that they will fill the 5 positions?
Here \(Y \sim HG(50, 10, 90)\) and so \[ P(Y \ge5) = 1- P(Y \le 4) \]
1-phyper(4, 10, 90, 50)## [1] 0.6297Note: the difference between the binomial and the hypergeometric distribution is that here we draw the balls without repetition. Of course, if n is small compared to N+M the probability of drawing the same ball twice is (almost) 0, so then the two distributions give the same answer.
using the binomial distribution for our example we would have found
1-pbinom(4, 50, 0.1)## [1] 0.5688quite different from the hypergeometric. On the other hand if our candidate pool had 1000 applicants, 100 of whom are suitable we would have found
1-phyper(4, 100, 900, 50)## [1] 0.5731A random variable X is said to have a Poisson distribution with rate \(\lambda\) (\(X \sim \text{Pois}(\lambda)\))if \[ P(X=k) = \frac{\lambda^x}{x!}e^{-\lambda}, k=0,1,2,.. \\ E[X] = \lambda \\ Var[X] = \lambda \]
One way to visualize the Poisson distribution is as follows say \(X \sim B(n,p)\) such that n is large and p is small. That is the number of trials is large but the success probability is small. Then X is approximately Poisson with rate \(\lambda = np\).
say you drive from Mayaguez to San Juan. Assume that the probability that on one kilometer of highway there is a police car checking the speed is 0.04. What is the probability that you will encounter at least 3 police cars on your drive?
If we assume that the police cars appear independently (?) then X = # of police cars \(\sim B(180, 0.04)\), so
\(P(X \ge 3 ) = 1 - P( X \le 2) =\)
1-pbinom(2,180,0.04)## [1] 0.9765One the other hand X is also approximately \(P(180*0.04)\) and so \(P(X \ge 3) =\)
1-ppois(2, 180*0.04)## [1] 0.9745