1. DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS
CHAPTER 3
DISCRETE RANDOM VARIABLES AND
PROBABILITY DISTRIBUTIONS

2. LEARNING OBJECTIVES Understand random variables
Calculate means and variances
Determine probabilities from cumulative distribution functions
Understand the assumptions of the discrete probability distributions
Calculate probabilities, determine means and variances for each of the discrete probability distributions

3. Concept of Random Variable
Summarize the outcome from a random experiment by a simple number
Used to describe the possible outcomes
Useful to associate a number with each outcome in the sample space
Variable that associates a number with the outcome of a random experiment is referred to as a random variable

4. Definition of Random Variables
Function that assigns a real number to each outcome in the sample space
Denoted by an uppercase letter such as X
Value of the random variable is denoted by a lowercase letter such as x

5. Two Types of Random Variables
Discrete and Continuous
Sometimes a measurement can assume any value in an interval of real numbers
Said to be a continuous random variable
Examples: Electrical current, length, pressure, temperature, time, voltage, weight
Sometimes the measurement is limited to integers
Said to be a discrete random variable
Examples: No. of scratches on a surface, No. of calls received per day, and No. of transmitted bits received in error

6. Probability Distribution
Describes probabilities associated with the possible values of X
Discrete Case
Specified by just a list of the possible values along with the probability of each
Continuous Case
Used to describe the probability distribution
Convenient to express the probability in terms of a formula

7. Discrete Probability Distribution
Definition
For a discrete random variable X with possible values x1, x2, x3, …, xn, the probability mass function (PMF) is
f(xi) = P(X = xi)
f(xi)  0 for all xi

8. Discrete Probability Distribution Examples
To better understand the PMF, consider
Example 1
Tossing a 6-sided die
f(x)=1/6 for X=1,2,3,4,5,6
Example 2
Check whether the following can serve as probability distribution
1.  f(x)= (x-2)/2 for x=1,2,3,4;
2.  h(x)= x2/25 for x=0,1,2,3,4;

9. Class Problem
The sample space of a random experiment is {a,b,c,d,e,f}, and each outcome is equally likely. A random variable is defined as follows:
Determine the probability mass function of X
f(0)=P(X=0)=1/6+1/6=1/3
f(1.5)=P(X=1.5)=1/6+1/6=1/3
f(3)=P(X=3)=1/6
Outcome a b c d e f x
1.5 2 3

10. Cumulative Distribution Function
Useful to provide cumulative probabilities such as P(Xx)
Cumulative distribution function (CDF) of a discrete random variable X, denoted as F(x), is
F(x) has the following properties:
F(x)= P(X  x)=
0  F(x)  1
If x  y, then F(x)  F(y)

11. Cumulative Distribution Function-Example
Determine the cumulative distribution of the random variable for which each outcome is equally likely in the following sample space:
Outcome a b c d e f x
1.5 2 3

12. Solution
Outcome a b c d e f x
1.5 2 3
f(x)
1/6

13. Calculating Probabilities from PMF and CDF
Interested to determine
Probabilities from cumulative distribution functions
Cumulative distribution functions from probability mass functions
Vise versa
Consider the following example

14. Example Consider the following PMF Determine the CDF f(y) y F(y) y y 1
0.4 y 1 2 3 4
f(y) .4 .3 .2 .1 y 1
F(y) y 1

15. Example Consider the following CDF
Note that Y can take on 1, 2, 3, or 4.
Determine the PMF
P(Y=1)= =0.4
P(Y=2)= =0.3, P(Y=3)= =0.2, P(Y=4)= =0.1
Hence, y 1 2 3 4
f(y) .4 .3 .2 .1

16. Mean of a Discrete Random Variable
PMF provides complete information about the properties of a random variable
Useful to have some summary measures of these properties
Mean or expectation
Denoted by E(X) or  represents an average value of the random variable
If a random variable takes on the values x1, x2,… , or xk with the probabilities f(x1), f(x2), …., and f(xk), its mean is
= E(X)= x1.f(x1) + x2.f(x2) xk.f(xk) Or
= E(X)=

17. Examples
Find the mean of the probability distribution of the number of heads obtained in three flips of a coin
Probabilities
1/8, 3/8, 3/8, and 1/8
= (0)1/8 + (1)3/8 + (2)3/8 + (3) 1/8 = 3/2
 Suppose Y is the total showing on a pair of dice, find the mean of the probability distribution
Calculated as

18. Variance of a Discrete Random Variable
Another important measure of the distribution of a random variable
Measures the spread or variability
Whereas mean measures central, the variance measures the deviation
Denoted as 2 or V(x)
2= V(X)=E(X-)2 =
Standard deviation of X is  =[V(X)]1/2

19. Examples
Find the variance of the probability distribution of the number of heads obtained in three flips of a coin
Variance
2= (0-3/2)2(1/8)+(1-3/2)2 (3/8)+(2-3/2)2(3/8)+(3- 3/2)2(1/8)=0.75
Find the variance of the total showing on a pair of dice
2 = V(X)= (2-7)2(1/36) + (3-7)2(2/36)+ … + (12-7)2(1/36) =

20. Class Problem
If the range of X is the set {0, 1, 2, 3, 4} and P(X=x)= 0.2, determine the mean and variance of the random variable
Solution
X can take values of 0,1, 2, 3 or 4
Summation of x f(x)
Hence, mean=2
Variance can be calculated as
V(X)= (0-2)2 (0.2)+(1-2)2 (0.2)+…=

21. Uniform Discrete Distribution
If X assumes the values x1,x2,….,xn with equal probability, then it has a discrete uniform distribution
Probability mass function
f(xi)= 1/n for xi= x1,x2,….,xn
Mean
E[X]=
Variance
V(X)=

22. Examples
When a light bulb is selected randomly from a box containing a red, a blue, a white, and a yellow bulb
Sample space is {red, blue, white, yellow} with probability 1/4
Hence, f(x)=1/4
When a die is tossed
Sample space is {1,2,3,4,5,6} with probability 1/6
f(x)=1/6
E[X]= (1*1/6+ 2*1/6+ 3*1/6+ 4*1/6+ 5*1/6+6*1/6)=3.5
V[X]=(1-3.5)2/6 + (2-3.5)2/6 + …+ (6-3.5)2/6 =35/12

23. Binomial Distribution
Two possible outcomes labeled success or failure
Referred to as a Bernoulli process
Conditions
Consists of n repeated trails
Results in two possible outcomes (Success or Failure)
Probability of success, denoted by p, remains constant
Trials are independent

24. Examples Tossing a coin 10 times
There are 10 trials, and they are identical
Each trial has only two outcomes
Probability of getting a H is 0.5 in each trial
Trials (tosses) are independent
In an operation, 5% of all machined parts are defective. 3 parts are randomly selected from the production line to determine if each of them is defective or good
Three identical trials
Each trial has two outcomes
p=0.05
Independent trials

25. PMF of the Binomial Distribution
Binomial distribution with parameters p and n=1,2,…
PMF
f(x)= px(1-p)n-x
Where x =0,1, 2,…,n.
E[X]= np
V (X)=np(1-p)=npq

26. Example
In an operation, 5% of all parts machined by a firm are defective
If three parts are randomly selected from the production line, what is the probability that exactly one of them will be defective?
Solution
n=3, x=1, and p=0.05
Substituting in the PMF
P(X=1)= f(1) = (1-0.05)3-1
P(X=1)=3*0.051(1-0.05)3-1 =

27. Class Problem
The random variable X has a binomial distribution with n=10 and p=0.01, determine the following probabilities
P(X=5)
= (0.01)5(1-0.01)10-5) =
P(X2)
=P(X=0)+P(X=1)+P(X=2)=
=0.99
P(3X<5)
=P(X=3)+ P(X=4)=1.14

28. Geometric Distribution
Closely related to the binomial experiment
Trials are conducted until a success is obtained
Let X denote the number of trials
PMF
P(X = x) = f(x) = (1-p)x-1p x = 1,2,3,…
Mean and Variance
=E(x)=1/p and 2=V(x)=(1-p)/p2

29. Example
The probability of a successful optical alignment in the assembly of an optical data storage product is 0.8
Assume the trials are independent
What is the probability the first successful alignment requires exactly four trials?
What is the probability that the first successful alignment requires at most four trials?
What is the probability that the first successful alignment requires at least four trials?

30. Example
Let X denote the number of trials to obtain in the first successful alignment.
Then X is a geometric random variable with p = 0.8
Solution
P(X = 4) = f(4) = 0.23(0.8) =
P(X  4) = P(X=1) + P(X = 2) + P(X =3) + P(X = 4)=0.9984
P(X  4) = 1  P(X < 4) = 1  = 0.008

31. Class Problem
Assume that each of your calls to a popular radio station has a probability of 0.02 of connecting, that is, of not obtaining a busy signal (calls are independent)
What is the probability that your first call that connects is your tenth call?
What is the probability that it requires more than five calls for you to connect?
What is the mean number of calls needed to connect?

32. Solutions
Let X denote the number of calls needed to obtain a connection
Then, X is a geometric random variable with p=0.02
P(X = 10) =
P(X>5) =
E(X) =1/0.02 = 50

33. Negative Binomial Distribution
Based on an experiment
Consists of a sequence of independent trials
Result in either a “S” or “F”
Probability of success is constant from trial to trial
Continues until a total of r success have been observed
PMF
P(X = x) = f(x) = (1-p)x-rpr
Mean and Variance
E[X] = r/p and V (X) = r(1-p)/p2

34. Example
Suppose that X is a negative binomial random variable with p=0.2 and r=4. Determine the following
E(X), P(X = 20), P(X = 19), P(X = 21)
Solution
E(X) = r/p =4/0.2 = 20
P(X = 20) = (1-p)x-rpr =
P(X = 19) =
P(X = 21) =

35. Hypergeometric Distribution
Assumptions
Population consists of N objects (finite)
Each classified as a “S” or “F” and K success
n individuals is selected without replacement
Random variable of interest is X, the number of successes in the sample

36. PMF of the Hypergeometric Distribution
X=number of successes in a sample of size n drawn from a population consisting of K successes and N-K failures
PMF is given
for x satisfying
max (0, n-N+K) x min (n,K)
Mean and Variance
E[X] = n and V (X) = n

37. Example A shipment of 20 machined parts contains 5 that are defective
If 10 of them are randomly chosen for inspection, what is the probability that 2 of the 10 will be defective?
Solution
x = 2, n = 10, k = 5, and N=20
f(x=2)= = 0.348

38. Class Problem
A lot of 75 washers contains 5 in which the variability in thickness around the circumference of the washer is unacceptable
A sample of 10 washers is selected at random, without replacement.
What is the probability that none of the unacceptable washers is in the sample?
What is the probability that at least one unacceptable washer is in the sample?
What is the probability that exactly one unacceptable washer is in the sample?
What is the mean number of unacceptable washers in the sample?

39. Solution
Let X denote the number of unacceptable washers in the sample of 10
P(X = 0)
P(X  1) =1- P(X=0)=0.5214
P(X = 1)=
E(X) =10*5/75=2/3

40. Poisson Distribution
Binomial, hypergeometric, and negative binomial distributions start with an experiment consisting of trials
Based on the number of outcomes occurring during a given time interval or in a specified regions
Examples
# of accidents that occur on a given highway during a 1-week period
# of customers coming to a bank during a 1-hour interval
# of TVs sold at a department store during a given week
# of breakdowns of a washing machine per month

41. Conditions
Consider the # of breakdowns of a washing machine per month example
Each breakdown is called an occurrence
Occurrences are random that they do not follow any pattern (unpredictable)
Occurrence is always considered with respect to an interval (one month)

42. The Probability Mass Distribution
X = number of counts in the interval
Poisson random variable with  > 0
PMF
f(x)= x=0,1,2,
Mean and Variance
E[X] =  , V (X) = 

43. Example
If a bank gets on average  = 6 bad checks per day, what are the probabilities that it will receive four bad checks on any given day?10 bad checks on any two consecutive days?
Solution
x = 4 and  = 6, then f(4) = = 0.135
 = 12 and x = 10, then f(10) = = 0.105

44. Class Problem
The number of failures of a testing instrument from contamination particle on the product is a Poisson random variable with a mean of 0.02 failure per hour.
What is the probability that the instrument does not fail in an 8-hour shift?
What is the probability of at least one failure in one 24-hour day?

45. Solution
Let X denote the failure in 8 hours. Then, X has a Poisson distribution with =0.16
P(X=0)=0.8521
Let Y denote the number of failure in 24 hours. Then, Y has a Poisson distribution with =0.48
P(Y1) = 1-P(Y = 0) =0.3812

46. The Poisson Approximation To The Binomial Distribution
When n is large and p is small, binomial probabilities are often approximates by
f(x)= for x = 0,1,2, … with  = np

47. Example
Assume 5% of the books at a certain bindery have defective bindings. Find the probability that 2 of 100 books bound by this bindery will have defective bindings, using
The binomial distribution
The Poisson distribution

48. Solution Using Binomial Using Poisson x = 2, n = 100, and p = 0.05
f(x) = (0.05)2(0.95)98 = 0.081
Using Poisson
x = 2,  = np = 100*0.05 = 5
f(2) = = 0.084

49. Next agenda
Continue our development of probability distributions with a discussion of several important continuous distributions