Consider the number of typos on each page of your textbook…

Or the number of car accidents between exit 168 and exit 178 on IN-65.

For binomial random variables, we know that they describe outcomes of binomial experiments.

Similarly, Poisson random variables describe outcomes of Poisson experiment

Properties of Poisson Experiment:◦ It measures the number of occurrence of an event

over an interval, area or space.

◦ The probability of an occurrence is the same for any two intervals of equal length/area.

◦ The occurrence or non-occurrence in any interval/area (area A) is independent of the occurrence or non-occurrence in any other interval/area (area B), if there is no overlap between A and B (or when A and B are disjoint)

• If a discrete random variable describes the outcome of Poisson experiment, we call this random variable a Poisson random variable.

• X~POI(λ)• P(X=k)=λ^k*e^(-λ) / k!– This is the probability of having k occurrences for a Poisson

random variable within a given interval/area.• If X is a Poisson random variable,– X must be non-negative integer– X has no upper limit

• There is only ONE parameter for Poisson distribution, λ.

• λ must be positive but does not have to be integer

If X~POI(λ), λ is both the mean AND variance of X. (Easier than binomial?)

I get 1.5 visit during my office hour on average. How likely will it be for me to have 3 visits during an office hour? (Suppose the number of visits follows a Poisson distribution)

What is the chance that I will have more than 5 visits in my next office hour?

On your textbook, there are 22 chapters and a total of 330 typos. You are reading chapter one and detected 20 typos, assuming each chapter of the textbook has the same number of pages, is that normal? (Suppose the number of typos follows a Poisson distribution).

You continued to read chapter two and also detected 20 typos. What is the probability that you get 20 typos on both chapter one and chapter two?

• A survey was conducted to research the number of car accidents on inter-state highway. It has found that there was an average of 2.5 accidents over each 100 miles of inter-state highway each year.

• a. It is about 100 miles from Lafayette to Edinburgh on IN-65, what is the probability that you see no accidents during one trip? (recall on your own experience.)

b. The Indiana department of transportation reported that during the year 2008, 4 accidents happened on IN-65 between Lafayette and Edinburgh, what is the probability for that?

Compare part a and b in example III, what can we say about using Poisson random variable to analyze real life events?

1. Suppose the number of typos in today’s Exponent follows a Poisson distribution with mean 18 and the number of typos in yesterday’s Exponent follows another Poisson distribution with mean 25, then the total number of typos in these two days’ Exponent still follows a Poisson distribution with mean 43 (=18+25)

Let Xf be the number of times you miss a class in fall semester and Xf~POI(35), let Xsp be the number of times you miss a class in spring semester and Xsp~POI(40), Xsu be the number of times you miss a class in the summer Xsu~POI(10),

Let Y be the total number of classes you miss for a school year, find the distribution of Y and the corresponding parameter.

Suppose we have more than one independent Poisson random variables, say n of them, X1, X2, …, Xn, each with parameter (λ1, λ2, …, λn etc). Here, independent means the corresponding region/interval for each of those variables are disjoint.

Then the sum of those variables X=X1+ X2+…+Xn, still follows a Poisson distribution with parameter (λ1+λ2+…+λn).

Sometimes, Poisson random variable (POI(λ))is used to approximate Binomial random variables (BIN(n,p)) when n is large and p is small.

In this case, we simply set λ=np.

Two empirical rules exist for using Poisson to approximate binomial:

1. n ≥ 20 and p ≤ 0.05 Or 2. n ≥ 100 and np ≤ 10

Suppose there is a weekly lottery which everyone has a 1/1000 chance of winning. What is the probability that you win this lottery five times a year.

A. Using binomial.

B. Using Poisson approximation.

Take a fair coin and toss it as many times as needed until you observe a head.

Let X= number of tosses that is needed. Sample points={H, TH, TTH, TTTH, …} Distribution of X

X 1 (H)

2(TH)

3(TTH)

4(TTTH)

K(TTT…TH)

P(X) 1/2 1/4 1/8 1/16 ?

Think about another example, if we keep tossing a biased coin with 70% of getting a tail and 30% of seeing a head, let X=number of tosses needed to get the first head.

Then the distribution of X is:

X 1 (H)

2(TH) 3(TTH) 4(TTTH) K(TTT…TH)

P(X)

0.3 0.7*0.3

0.7*0.7*0.3

0.7*0.7*0.7*0.3

0.7*0.7*…*0.7*0.3

If we repeat an experiment with two outcomes, success/failure, with probability p for success and q=1-p for failure, the number of trials needed to get the first success follows a Geometric distribution, say,

X~Geo(p). P(X=k)=(1-p)^(k-1)*p

• Similarities and differences between Binomial and Geometric random variable.– Similarities: independent trials of the identical

experiment, probabilities of success/failure consistent.

–Differences: • For Binomial, we know how many trials we have in

total.• For Geometric, we don’t know it, actually that

number is not of interest.

Someone is trying to take the road test to get a driver’s license. If the probability of passing the test is 40%, what is the probability that this person will pass the test at second shot?

What is the probability that someone will pass the road test in 5 trials?

Given that someone has taken the test 4 times and still has not got the license, what is that person’s chance of passing it the next time?

If X~Geo(p), then ◦E(X)=1/p◦Var(X)=(1-p)/(p^2)

On average, how many times does one have to take the test to get the driver’s license?