Download Basic Probability Rules and Binomial Distribution

Basic Probability Rules and Binomial Distribution
 Definitions
 Experiment

Event

Simple Event

Sample Space
 Probability Rules
 Simple

Intersection or Joint

Union

Complement

Conditional

Special Cases
 Binomial
Purpose: Determining probabilities based on counting techniques and formulas. Probability is
needed for the third Building Block (To evaluate an error use probability) and fourth Building
Block (A margin of error is the largest error expected with a specified probability).
Note the percent of the population that takes on a specific value is equivalent to a probability
only if the item is randomly selected.
1. Definitions
1.1 Experiment – a repeatable process that has an outcome that is recorded
Examples:
 You flip a coin.
 I record the grades of students.
 You ask a politician two questions: What is your party affiliation? How do you plan to
vote on the next bill?
1.2 Event – an outcome of the event.
 The politician says that she is a Republican.
 The tossed coin is a head.
 The grade is a 90.
1.3 Simple Events – Events that cannot be decomposed into other events.
 The tossed coin is a head.
 The politician says that she is Republican but that can be decomposed into (a) a
Republican that says yes and (b) a Republican that says no. The answer to the two
questions is a simple event.
1.4 Sample Space – the list of all possible simple events or outcomes.
For the coin the sample space is H or T
For the politician the sample space is (a) Republican that votes yes, (b) Republican that votes no,
(c) Democrat that votes Yes, etc.
2. Probability Rules
Example Pivot Table: Using the following file:
http://wweb.uta.edu/faculty/eakin/busa5325/ContTableExample.xls we can create a pivot table
examining the answers of 100 politicians to two questions (a) “What is your party affiliation?”
and (b) “How do you plan to vote on the next bill?”
Vote Yes
Vote No
Total
Democrat
34
6
40
Republican
18
22
40
Other
8
12
20
Total
60
40
100
Assuming any one of the 100 politicians is as likely to be chosen as any other. (What Building
Block is this?)
2.1 Events (also called Marginal Events)
(Marginal) Probability of choosing a Democrat
(Marginal) Probability of choosing someone who voted Yes
2.2 Intersection or Joint Events (These are also simple events)
Probability of choosing a Democrat and who voted Yes
2.3 Complementary Events
Probability of choosing not choosing a Democrat?
Formula: Pr(D’) = Pr(not D) = 1- Pr(D)
2.4 Union of Events
Probability of choosing a Democrat or someone who voted Yes
(Addition Rule) Formula: Pr(D or Y) = Pr(D) + Pr(Y) – Pr(D and Y)
2.5 Conditional Events
Probability of choosing a Democrat from among those who voted Yes
Conditional Probability: Pr(D|Y) = Pr(D and Y) / Pr(Y)
rewriting this leads to the
(Multiplicative Rule) Formula: Pr(D and Y) = Pr(Y)Pr(D|Y)
2.6 Examples– For more examples control-click on the link below to load an Excel file. To
create new exercises, press the F9 key. The answers are below the problem. Be sure to turn off
auto-recalculation or new examples may be accidently constructed erasing the one you are on.
http://wweb.uta.edu/faculty/eakin/busa3321/creatingprob.xls
Exercise: Using Internet Explorer answer the questions on the following web page. If you get a
question wrong, you can try it again. When you get all the answers correct, copy the screen to
Word and post on Blackboard.
https://wweb.uta.edu/faculty/eakin/asps/Examples/ProbQues.asp
2.7 Special Cases:
2.71. Mutually Exclusive Events:
Probability of choosing a person who voted Yes and voted No.
Formula: Pr (N and Y) = 0
Which leads to a special case of the additive rule:
Pr ( N or Y) = Pr(N) + Pr(Y)
2.7.2 Independent Events
Vote Yes
Vote No
Total
Prob(Yes)
Pr(Yes | Democrat)
Pr(Yes | Republican)
Democrat
24
16
40
Republican
24
16
40
Other
12
8
20
Total
60
40
100
Pr(Yes | Other)
Pr(Yes and Democrat)
Formula: Pr(Y|D) = Pr(Y) and changes the multiplicative rule to
Pr(D and Y) = Pr(D)Pr(Y)
2.8 Conditional Probability Questions
2.8.1 Example: Examples– For more examples control-click on the link below to load an Excel
file. To create new exercises, press the F9 key. The answers are below the problem. Be sure to
turn off auto-recalculation or new examples may be accidently constructed erasing the one you
are on.
http://wweb.uta.edu/faculty/eakin/busa3321/bayesexamples.xls
Exercise: Using Internet Explorer answer the questions on the following web page. If you
get a question wrong, you can try it again. When you get all the answers correct, copy the screen
to Word and post on Blackboard.
http://wweb.uta.edu/faculty/eakin/asps/examples/Bayesques.asp
3. Binomial Probability Rules
3.1 Situation: A random sample of size n is taken from a population of values that take on only
two values. One value is declared to be a success and the other a failure. The proportion of
values in the population that are successes has the symbol, p
Binomial Random Variable:
 X is the Number of Successes in the Sample
 X = 0, 1, …. , n
Population: Mean and Standard Deviation of X
  = n p
   np(1  p)
3.2 Finding Probabilities
3.2.1 Formula (useful for finding the probability that X takes on a particular value, x):
prob(X  x) 
n!
p x (1  p) n  x
x!(n  x )!
The first part of the formula is the number of ways, x successes can be found in n trials
n!
x!(n  x )!
Note that n! means to take a number, n, and multiply it by every positive integer smaller than it.
5! = 5*4*3*2*1 = 120. A special case is 0! = 1. The development of this formula is shown in
Section 2.3 below.
Example: Find Pr(x=3 | n=4 and p =0.25) =
4!
0.253 (1  .25) 43  4 * 3 * 2 *1 0.015625(0.75)
3!(4  3)!
3 * 2 *1 * (1)
 4 * 0.015625(0.75)  0.046875
3.2.2 Tables: The column values corresponds to values of p and the rows to n and x
2.2.2.1 Pr(X= a value) use http://wweb.uta.edu/faculty/eakin/busa3321/BinomialTable.doc
Example: Find Pr(x=3 | n=4 and p =0.25) = 0.047. Excerpt from table:
3.2.2.1 Pr(X≤ a value) use
http://wweb.uta.edu/faculty/eakin/busa3321/BinomialCumulativeProbability.doc
Example Find Pr(x≤ 2 | n=5 and p= 0.30) = 0.837
3.2.3 Empirical Rule
Requirement to use empirical rule: (both np and n(1-p) must be greater than or equal to
5)
Example: n = 100, p = 0.50
  =np = 100(0.50) = 50
   np (1  p)  100 (0.5)(1  .05)  5
 Since np = n(1-p) = 50 > 5 we can use the empirical rule
Pr (40 ≤ X ≤ 45) is approximately 13.5%
.
3.3 Developing the binomial formula by example (This will not be tested on an exam)
We will be using the special cases of section 1.6
If A and B are mutually exclusive then Pr(A or B) = P(A) + P(B).
if A and B are independent then Pr(A and B) = P(A)P(B)
Consider a case where there is a 5 question multiple choice exam with four possible
answers on each question and each question is independent of the others. Suppose
you decide to guess at each question.
If you guess, then p, the probability of a success in any one question, = ¼ = 0.25 and
probability of getting a question wrong is 3/4 = 0.75
Notation let I1 stand for getting question 1 incorrect and C1 stand for getting
question 1 correct.
3.3.1 First find the probability of getting no questions correct = Pr(X=0).
Because of independent questions Pr (question 1 incorrect and question 2
incorrect and 3 incorrect and 4 incorrect and 5 incorrect)
Pr(X=0) = (0.75)*(0.75) *(0.75) *(0.75) *(0.75) = (0.75)5= 0.2373046875
3.3.2 Next find the probability of getting at one question correct Pr(X=1). There are
five ways this could occur. Getting question 1 correct and not the others or getting
question 2 correct and not the others, etc.
Pr (getting question 1 correct and not the others)
= (.25)*(0.75) *(0.75) *(0.75) *(0.75)
= (0.25)(0.75)4
= 0.0791015625
Pr(getting question 2 correct and not the others)
= (0.75)*(0.25)*(0.75)*(0.75)*(0.75)
= (0.25)(0.75)4
= 0.0791015625
Same goes for the other three ways. Since the five ways are mutually exclusive,
we can add the probabilities together or take any one of them and multiply by 5
Pr(X=1) = 5 (0.25)(0.75)4 = 0.3955078125
3.3.3. The probability of two successes out of five could occur ten different ways.
[Notation let I1 stand for getting question 1 incorrect and C1 stand for getting
question 1 correct. ]
(C1 and C2 and I3 and I4 and I5) or
(C1 and I2 and C3 and I4 and I5) or
(C1 and I2 and I3 and C4 and I5) or
(C1 and I2 and I3 and I4 and C5) or
(I1 and C2 and C3 and I4 and I5) or
(I1 and C2 and I3 and C4 and I5) or
(I1 and C2 and I3 and I4 and C5) or
(I1 and I2 and C3 and C4 and i5) or
(I1 and I2 and C3 and I4 and C5) or
(I1 and I2 and I3 and C4 and C5)
Note that the value 10 can be found by substituting n=5 and x =2 into the
following formula.
However in each case the probability is (0.25)2(0.75)3 which is the probability of a
success raised to the number of successes and the probability of a failure raised to
the number of failures.
Since each of the ten cases are mutually exclusive,
Pr (X=2) = 10 (0.25)2(0.75)3=0.263671875
Which is the formula
with n=5, X=2 and  = 0.25
he other cases pr(x=3), Pr(X=4) and Pr(X=5) can be developed similarly.
Compare these probabilities to the ones in the tables.