1. Chapter 3: Random Variables and Probability Distributions
Definition and nomenclature
A random variable is a function that associates a real number with each element in the sample space.
We use a capital letter such as X to denote the random variable.
We use the small letter such as x for one of its values.
Example: Consider a random variable Y which takes on all values y for which y > 5.
P(X > 5)
P(Y < 3)
a random variable, X, is a function that associates a real number with each element in the sample space and x is one of the values that X can take.
JMB Chapter 3 Lecture 1
EGR Spring 2011

2. Defining Probabilities: Random Variables
Examples:
Out of 100 heart catheterization procedures performed at a local hospital each year, the probability that more than five of them will result in complications is
P(X > 5)
Drywall anchors are sold in packs of 50 at the local hardware store. The probability that no more than 3 will be defective is
P(Y < 3)
P(X > 5)
P(Y < 3)
a random variable, X, is a function that associates a real number with each element in the sample space and x is one of the values that X can take.
JMB Chapter 3 Lecture 1
EGR Spring 2011

3. Discrete Random Variables
Problem Page 55 Modified
Assume someone spends $75 to buy 3 envelopes. The sample space describing the presence of $10 bills (H) vs bills that are not $10 (N) is:
S = {NNN, NNH, NHN, HNN, NHH, HNH, HHN, HHH}
The random variable associated with this situation, X, reflects the outcome of the experiment
X is the number of envelopes that contain $10
X = {0, 1, 2, 3}
Note: if the number of possible solutions is countable, the variable is discrete
S = {NNN, NNH, NHN, HNN, NHH, HNH, HHN, HHH}
X = {0, 1, 2, 3}
the probability distribution function (x, f(x)) – see definition 3.4, pg. 66
JMB Chapter 3 Lecture 1
EGR Spring 2011

4. Discrete Probability Distributions 1
The probability that the envelope contains a $10 bill is 275/500 or .55
What is the probability that there are no $10 bills in the group?
P(X = 0) =(1-0.55) * (1-0.55) *(1-0.55) =
P(X = 1) = 3 * (0.55)*(1-0.55)* (1-0.55) =
Why 3 for the X = 1 case?
Three items in the sample space for X = 1
NNH NHN HNN
P(X=0) = P(not in the 1st envelope ∩ not in the 2nd ∩ not in the 3rd) = (1-275/500)3 = (0.45)3 =
P(0) =(1-0.55)^3 =
P(1) =3*((0.55)*(1-0.55)^2) =
P(2) =3*(0.55^2*(1-0.55)) =
P(3) = 0.55^3 =
(students fill in the table)
JMB Chapter 3 Lecture 1
EGR Spring 2011

5. Discrete Probability Distributions 2
P(X = 0) =(1-0.55) * (1-0.55) *(1-0.55) =
P(X = 1) = 3*(0.55)*(1-0.55)* (1-0.55) =
P(X = 2) = 3*(0.55^2*(1-0.55)) =
P(X = 3) = 0.55^3 =
The probability distribution associated with the number of $10 bills is given by:
P(X=0) = P(not in the 1st envelope ∩ not in the 2nd ∩ not in the 3rd) = (1-275/500)3 = (0.45)3 =
P(0) =(1-0.55)^3 =
P(1) =3*((0.55)*(1-0.55)^2) =
P(2) =3*(0.55^2*(1-0.55)) =
P(3) = 0.55^3 =
(students fill in the table) x 1 2 3
P(X = x)
JMB Chapter 3 Lecture 1
EGR Spring 2011

6. Example 3.8, pg 80 Shipment: 8 computers of which 3 are defective
Random purchase of 2 computers
What is the probability distribution for the random variable X = defective computers purchased?
Possibilities: X = 0 X =1 X = 2
Let’s start with P(X=0) P = specified target / all possible
(0 defectives and 2 nondefectives are selected)
(all ways to get 0 out of 3 defectives) ∩ (all ways to get 2 out of 5 nondefectives) (all ways to choose 2 out of 8 computers) (all ways to choose 2 out of 8 computers)
P(X = 0) = P(0 defectives and 2 nondefective) =
(all ways to get 0 out of 3 defectives) ∩ (all ways to get 2 out of 5 nondefective)
(all ways to choose 2 out of 8 computers) (all ways to choose 2 out of 8 computers)
JMB Chapter 3 Lecture 1
EGR Spring 2011

7. Discrete Probability Distributions
The discrete probability distribution function (pdf)
f(x) = P(X = x) ≥ 0
Σx f(x) = 1
The cumulative distribution, F(x)
F(x) = P(X ≤ x) = Σt ≤ x f(t)
Note the importance of case: F not same as f
JMB Chapter 3 Lecture 1
EGR Spring 2011

8. Probability Distributions
From our example, the probability that no more than 2 of the envelopes contain $10 bills is
P(X ≤ 2) = F (2) = _________________
F(2) = f(0) + f(1) + f(2) =
 (OR 1 - f(3))
The probability that no fewer than 2 envelopes contain $10 bills is
P(X ≥ 2) = 1 - P(X ≤ 1) = 1 – F (1) = ________
1 – F(1) = 1 – (f(0) + f(1)) = = .575
 (OR f(2) + f(3))
F(2) = f(0) + f(1) + f(2) =  (OR 1 - f(3))
1 – F(1) = 1 – (f(0) + f(1)) = =  (OR f(2) + f(3))
JMB Chapter 3 Lecture 1
EGR Spring 2011

9. Another View The probability histogram JMB Chapter 3 Lecture 1
EGR Spring 2011

10. Your Turn …
The output of the same type of circuit board from two assembly lines is mixed into one storage tray. In a tray of 10 circuit boards, 6 are from line A and 4 from line B. If the inspector chooses 2 boards from the tray, show the probability distribution function associated with the selected boards being from line A. x
P(x) 1 2
P(x = 0) =
JMB Chapter 3 Lecture 1
EGR Spring 2011

11. Continuous Probability Distributions
In general,
The probability that the average daily temperature in Georgia during the month of August falls between 90 and 95 degrees is
The probability that a given part will fail before 1000 hours of use is
Probability density function f(x)
JMB Chapter 3 Lecture 1
EGR Spring 2011

12. Visualizing Continuous Distributions
The probability that the average daily temperature in Georgia during the month of August falls between 90 and 95 degrees is
The probability that a given part will fail before 1000 hours of use is
JMB Chapter 3 Lecture 1
EGR Spring 2011

13. Continuous Probability Calculations
The continuous probability density function (pdf)
f(x) ≥ 0, for all x ∈ R
The cumulative distribution, F(x)
Example: the uniform distribution (i.e., f(x) = 1, 1 < x < 2)
1. what is the area of the rectangle? (1) The total area under the curve is P(S) and so will always be 1.
JMB Chapter 3 Lecture 1
EGR Spring 2011

14. Example: Problem 3.7, pg. 88
The total number of hours, measured in units of 100 hours
x, 0 < x < 1
f(x) = 2-x, 1 ≤ x < 2
0, elsewhere
P(X < 120 hours) = P(X < 1.2)
= P(X < 1) + P (1 < X < 1.2)
NOTE: You will need to integrate two different functions over two different ranges.
b) P(50 hours < X < 100 hours) =
Which function(s) will be used? {
P(X < 1.2) = P(X < 1) + P (1 < X < 1.2) = ∫01xdx + ∫11.2 (2-x)dx = (x2/2)|01 + (2x- x2/2)|11.2 =0.68
P(.5 < X < 1) =
JMB Chapter 3 Lecture 1
EGR Spring 2011