1. IE 360: Design and Control of Industrial Systems I
Lecture 12
More Specific
Discrete Random Variables
IE 360: Design and Control of Industrial Systems I
References
Montgomery and Runger
3-7, 3-8, 3-9
Copyright  2010 by Joel Greenstein

2. Geometric RV Here is the story for a geometric rv
To be more concrete, think of flipping a coin that has 30% chance of being heads until we get the first head. The number of flips we need is a geometric rv.
Here is the story for a geometric rv
You do NOT have a fixed number of trials
Instead, you want to perform identical,
independent Bernoulli trials until one success occurs. Then we stop
Each trial is the same, in the sense that each trial can result in a success or a failure with a constant probability p of being a success AND these trials are independent
If the random variable X takes on values 1, 2, 3, … (doesn’t stop) where each value is the number of trials until the first success occurs, then X is a geometric rv with pmf given by
We use the format “f(x; p)” to emphasize we only need to know the probability of success in each trial
X~Geo(p) indicates that X is distributed as a geometric rv with the same probability of success, p, on each of the Bernoulli trials.

3. Geometric RV: common values to compute
The pmf is derived with the following logic
The probability of x-1 independent trials with no success occurring is qx-1
The probability of 1 trial (independent of the others) being a success is p
Expected value of a geometric rv
Presented without proof
Variance of a geometric rv
Trial 1
Success; X=1
Failure
Trial 2
Success; X=2
Failure
Trial 3
Success; X=3
Failure, keep going ….

4. Geometric RV Example
The probability that a student pilot passes the written test for a private pilot’s license is 0.7 each time he or she takes the test. Find the probability that the student will pass the test on the third try.
Step 1: Ensure this is really a geometric rv problem.
We want to know how many attempts the student will need. And we are given that each attempt is independent and identically distributed (the probability of success is the same in all trials). Let X be the number of tries the student needs. Then X is a geometric rv.
Step 2: Identify the parameter
For a geometric rv, we need p
p=0.7
Step 3: Identify the question being asked
P(X=3)
Step 4: Answer it
Now: what is the probability that the student pilot needs 3 or more attempts?
P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5) + ….. + P(X = 70000) + ….. (yuck)
Instead, use P(X ≥ 3) = 1 – P(X < 3) = 1 – P(X = 1) – P(X = 2) = 0.09

5. Forgetfulness Property
What if I told you that the student pilot has definitely not passed the written exam on the first 3 attempts, and then ask, given this knowledge, “What is the probability that the student pilot requires 4 or more additional attempts?
Now, we need to use some conditional probability
We want P(X ≥ 7|X ≥ 4)
We know that the student pilot will need four or more attempts, since we are told that the student pilot has not passed on the first 3 attempts.
Let’s apply the definition of conditional probability
What is P(X ≥ 7 ∩ X ≥ 4)? It is the probability that the student pilot needs at least 7 attempts AND needs at least 4 attempts. But really, if the student needs at least 7 attempts, they automatically need at least 4 attempts (because 7 is more than 4).
Let’s compute this in that case.

6. Forgetfulness Property – Crazy Math
This is crazy math … you can just jump to the end …
The long and the short of it is this: P(X≥7 | X≥4) = P(X≥4)

7. Forgetfulness Property – In summary
The forgetfulness property in general is
The geometric distribution has this property
It is also called the lack of memory property
This is important!!!!! Especially for IE 381 and Markov chains
It means that for a geometrically distributed random variable, the “future” (i.e., X ≥ s) does not depend on the past (i.e., X < t)

8. Poisson RV The Poisson distribution is used a lot in IE 381 and IE 482
Useful for situations where the event of interest has a low probability of occurrence but many opportunities for occurrence
Defects in paint, potholes in a road, typos in a book
It is a count of something, so it is definitely a discrete rv
It can be used to approximate the binomial distribution when n is large and p is small
The Poisson distributed random variable X takes on values 0, 1, 2, 3, … (doesn’t stop) with pmf given by
The parameter λ is the average number of items over some “area” or “volume” or “time”
The average number of defects (count) in a specified number of square inches of paint (area)
The average number of customers arriving (count) in a specified number of minutes (time)
X~Pois (λ) indicates that X is distributed as a Poisson rv with parameter λ

9. Poisson RV: Common values to compute
Expected value of a Poisson rv
Presented without proof
Variance of a Poisson rv
Basic probability
Consider R, the number of defects in a process, distributed as a Poisson rv with an average number of 80 defects per hour
Find the probability of 36 defects being found in ½ hour.
In this case, So
The Poisson rv is critically important for IE 381, especially for queuing theory (the mathematical study of lines and arrivals and service)

10. Poisson rv Example
A store sees, on average, 5 customers enter per hour. What is the probability that no customers enter during a specific 4 hour period of time?
Step 1: Ensure this is really a Poisson rv problem.
It isn’t anything else we know. And, we are told a rate as the average value of some variable. Best assumption is that it is Poisson. Let X be the number of customers that enter in a 4 hour period.
Step 2: Identify the parameters
For a Poisson rv, we need λ
We have 5 customers entering per hour.
So in 4 hours, we expect
Step 3: Identify the question being asked
P(X=0)
Step 4: Answer it
Doesn’t it make sense that this answer is really small?

11. Poisson rv as an approximation to the binomial
If n is “very large” and p is “very small”, then we can use the Poisson as an approximation of the binomial
What does “very large” and “very small” mean?
If n>20 and p<0.05, we can use the approximation
If n>100 and np<10, we can definitely use the approximation
Why bother?
The binomial pmf can be “hard” to calculate for large n because the factorial of a large number (like 5000) is hard to calculate
In the “old days” (1800s) this was a big problem, which motivated the development of the Poisson rv in the first place
How to do the approximation
Assume X is distributed as a binomial rv with n and p. If n is “very large” and p is “very small”, then use the following

12. Last example
A mutation occurs in 1 of every 10,000 births. If 20,000 babies are born, what is the probability that at least one baby has the mutation? Use the binomial and the Poisson approximation to the binomial to answer this question.
Step 1: Identify the RV
It was given as a binomial, so let M be the number of mutations in 20,000 babies
Step 2: Identify parameters
The binomial requires n and p. We are given n=20,000 and p=1/10000
The Poisson approximation to the binomial says to use np as λ. Therefore, we will use λ = 2
Step 3: Set up the question
In either case, we need P(M ≥ 1) = 1- P(M = 0)
Same to 4 decimal places

13. Related reading Montgomery and Runger, Sections 3-7, 3-8, and 3-9
The assigned reading also covers the negative binomial and the hypergeometric distributions, which were not covered in these PowerPoints. Study these in the reading.
Don’t forget the forgetfulness property of the geometric distribution. You won’t find it in the textbook.
Some links that might help
Keep this one – it is really good!!!! It will probably even help in IE 381
Now you are ready to do HW12