4.3 - Triola textbook 5.3 - Sullivan textbook

1. Elementary Statistics
Chapter 4:
Probability
4.3 Complements and
Conditional
Probability, and Bayes’
Theorem
1

2. Chapter 4: Probability
4.1 Basic Concepts of Probability
4.2 Addition Rule and Multiplication Rule
4.3 Complements and Conditional Probability, and Bayes’ Theorem
4.4 Counting
4.5 Probabilities Through Simulations (available online)
2
Objectives:
• Determine sample spaces and find the probability of an event, using classical probability or empirical probability.
• Find the probability of compound events, using the addition rules.
• Find the probability of compound events, using the multiplication rules.
• Find the conditional probability of an event.
• Find the total number of outcomes in a sequence of events, using the fundamental counting rule.
• Find the number of ways that r objects can be selected from n objects, using the permutation rule.
• Find the number of ways that r objects can be selected from n objects without regard to order, using the combination rule.
• Find the probability of an event, using the counting rules.

3. Notation: P(A) = the probability of event A , 0 ≤ P(A) ≤ 1
The sum of the probabilities of all the outcomes in the sample space is 1: 𝑃𝑖 = 1
Impossible Set: If an event E cannot occur (i.e., the event contains no members in the sample space), its probability
is 0. P(A) = 0
Sure (Certain) Set: If an event E is certain, then the probability of E is 1. P(E) = 1
Complementary Events: 𝐴: consists of all outcomes that are not included in the outcomes of event A.
𝑃 𝐴 = 1 − 𝑃(𝐴)
The actual odds against event A: O 𝐴 =
𝑃 𝐴
𝑃(𝐴)
, (expressed as form of a:b or “a to b”); a and b are integers.
The actual odds in favor of event A: O 𝐴 =
𝑃 𝐴
𝑃 𝐴
(the reciprocal of the actual odds against the even) If the odds
against A are a:b, then the odds in favor of A are b:a.
The payoff odds against event A: Payoff odds against event A = (net profit):(amount bet)
Payoff odds against event A =
net profit
amount bet
net profit = (Payoff odds against event A)(amount bet)
Value: Give the exact fraction or decimal or round off final decimal results to 3 significant digits. 3
Recall 4.1 Basic Concepts of Probability

4. Addition Rule:
P(A or B) = P(in a single trial, event A occurs or event B occurs or they both occur)
P(A or B) = P(A) + P(B) − P(A and B)
where P(A and B) denotes the probability that A and B both occur at the same time as an
outcome in a trial of a procedure.
Multiplication rule:
A tool to find P(A and B), which is the probability that event A occurs and event B occurs.
The word “and” in the multiplication rule is associated with the multiplication of
probabilities.
Sampling with replacement: Selections are independent events. P(A ∩ B) = P(A)  P(B)
Sampling without replacement: Selections are dependent events. P(A and B) = P(A)  P(B |
A)
Disjoint (or mutually exclusive): Events A and B are disjoint (or mutually exclusive) if
they cannot occur at the same time. (That is, disjoint events do not overlap.)
Complementary Events: Rules 𝑷 𝑨 + 𝑷 𝑨 = 𝟏
𝐴 𝑎𝑛𝑑 𝐴 must be disjoint.
Recall 4.2 Addition & Multiplication Rules
Venn
Diagram
for
Events
That Are
Not
Disjoint
Venn
Diagram
for
Disjoint
Events
4

5. Key Concept: We extend the use of the multiplication rule to include the
probability that among several trials, we get at least one of some specified
event. We consider conditional probability: the probability of an event
occurring when we have additional information that some other event has
already occurred. We provide a brief introduction to Bayes’ theorem.
4.3 Complements and Conditional Probability, and Bayes’Theorem
Complements: The Probability of “At Least One”
When finding the probability of some event occurring “at least once,” we
should understand the following:
• “At least one” has the same meaning as “one or more.”
• The complement of getting “at least one” particular event is that you get no
occurrences of that event.
P(at least one occurrence of event A) = 1 − P(no occurrences of event A)
5( t least 1) 1 ( )P A P None 

6. Example 1
If a couple plans to have 3 children, (Assume equally likely and independency)
a) what is the probability that they will have at least one girl?
b) What is the probability that they will have at most one girl?
6
Complementary Events: Rules ( ) 1 ( )P A P A 
Solution: Two ways: Write the sample space
P( 𝐴 ) = P(All boys)=
1
2
3
=
1
8
1) {ggg, ggb, gbg, gbb, bgg, bgb, bbg, bbb}, P (At least one girl) = 7/8
Interpretation There is a 7/8 probability that if a couple has 3 children,
at least 1 of them is a girl.
2) (At Least One) 1 ( )P P None 
3
1 1 7
(At Least One Girl) 1 1
2 8 8
P
 
     
 
1 3 4 1
8 8 8 2
   ) (At Most One Girl) (0) (1)b P P P 
( t least 1) 1 ( )P A P None 

7. Example 2
Assume that 3% of ties sold in the United States are bow ties (B). If 4
customers who purchased a tie are randomly selected, find the probability
that at least 1 purchased a bow tie.
7
Complementary Events: Rules ( ) 1 ( )P A P A 
   : 0.03, 1 0.03 0.97, 4Given P B P B n    
         no bow tiesP P B P B P B P B   
   at least 1 bow tie 1 no bow tiesP P 
 
4
0.97 0.885 
1 0.885 0.115  
( t least 1) 1 ( )P A P None 
 
4
1 0.97 

8. Example 3
Assume 6% of damaged iPads are damaged by backpacks.
If 20 damaged iPads are randomly selected, find the probability of getting at least one
that was damaged in a backpack. Is the probability high enough so that we can be
reasonably sure of getting at least one iPad damaged in a backpack?
8
Complementary Events: Rules ( ) 1 ( )P A P A 
   : 0.06, 1 0.06 0.94, 20Givven P D P D n    
   
20
no Damage by backpackP P D 
 
 
 
at least 1 Damaged by Bagback
1 None Damaged
P
P 
 
20
0.94 0.2901 
1 0.2901 0.7099  
Interpretation:
n = 20 & P = 0.7099 probability of getting at least 1 iPad damaged in a backpack.
This probability is not very high, so to be reasonably sure of getting at least 1 iPad damaged in a backpack, we should select more than
20 damaged iPads.
Very High Probability means: As n (the sample size) gets larger the probability should get closer to 1.
( t least 1) 1 ( )P A P None 

9. A conditional probability of an event is a probability obtained with the additional
information that some other event has already occurred.
Notation
P(B | A) denotes the conditional probability of event B occurring, given that event A
has already occurred.
It can be found by assuming that event A has occurred and then calculating the
probability that event B will occur.
9
4.3 Complements and Conditional Probability, and Bayes’Theorem
 
 
 
 
 
 
Conditional Probability
and
and
P A B
P A B
P B
P A B
P B A
P A


( )
Prob of A Given B: ( ) , ( ) 0
( )
( )
Prob of B Given A: ( ) ; ( ) 0
( )
P A B
P A B P B
P B
P A B
P B A P A
P A

 

 

10. Example 4
The probability that Sean parks in a no-parking zone and gets a parking
ticket is 0.08, and the probability that Sean cannot find a legal parking
space and has to park in the no-parking zone is 0.40. If Sean arrives at
school and has to park in a no-parking zone, find the probability that he
will get a parking ticket.
10
Conditional Probability
Solution: N = parking in a no-parking zone & T = getting a ticket
 
 
 
|
P N T
P T N
P N


0.08
0.20
0.40
 
   : and 0.08, 0.40Given P N T P N 
 
 
 
 
 
 
Conditional Probability
and
and
P A B
P A B
P B
P A B
P B A
P A



11. Example 5
A group of 150 randomly selected CEO’s was tested for
personality type. The following table gives the result of this
survey. (Contingency table or frequency distribution table
for Bivariate data)
a. If one CEO is selected at random from this group, find
the probability that this CEO
i. Has a Type A personality.
ii. Is a Woman.
iii. Is a Man given that he has a Type A personality.
iv. Has a Type B personality given that she is a Woman.
v. Has Type A personality and is a Woman.
11
blank Type A (A) Type B (B) Total
Men (M) 78 42 120
Women (W) 19 11 30
Total 97 53 150
 
 
 
|
P A B
P A B
P B


 
 
( )
n A
P A
n s
 78 19 97
150 150

 
 
 
( )
n w
P w
n s
 19 11 30
150 150

 
1
5

 
 
 
|
P M A
P M A
P A


78
97

 
 
 
|
P B W
P B W
P W


11/150 11
30/150 30
 
 
 
( )
n A W
P A W
n s

 
19
150


12. Example 5 Continued
A group of 150 randomly selected CEO’s was tested
for personality type. The following table gives the
result of this survey. (Contingency table or
frequency distribution table for Bivariate data)
a. If one CEO is selected at random from this group,
find the probability that this CEO
i. Is a Man or has a Type B personality
b. Are the events Woman and Type A personality
mutually exclusive? Are the events Type A and
Type B personalities mutually exclusive?
c. Are the events Type A personality and Men
independent?
12
 
 
 
|
P A B
P A B
P B


P(M or B) = P ( M ) + P ( B ) - P (M and B)
: ( ) ( | ) ( ) ( ) ( | )OR P M P B M P M B P B P M B  
120 53 120 42 120 53 53 42 131
150 150 150 120 150 150 150 53 150
       
19
( ) ,
150
P A w No  ( ) 0,P A B Yes 
 
120 78
( ) | ,
150 120
P M P M A NO  
blank Type A (A) Type B (B) Total
Men (M) 78 42 120
Women (W) 19 11 30
Total 97 53 150
( ) ( ) ( )
( )
( )
n M n B n M B
P M B
N s
 

120 53 42 131
150 150
 
 

13. Example 6: Find the following using the table:
a. If 1 of the 555 test subjects is randomly selected,
find the probability that the subject had a positive
test result, given that the subject actually uses
drugs. That is, find P(positive test result | subject
uses drugs).
b. If 1 of the 555 test subjects is randomly selected,
find the probability that the subject actually uses
drugs, given that he or she had a positive test result.
That is, find P(subject uses drugs | positive test
result).
13
blank Positive Test Result (P)
(Test shows drug use.)
Negative Test Result (N)
(Test shows no drug use.)
Total
Subject Uses
Drugs (D)
45
(True Positive)
5
(False Negative)
50
Subject Does Not
Use Drugs (ND)
25
(False Positive)
480
(True Negative)
505
Total 70 485 555
 
 
 
|
P P D
P P D
P D


45/555
50/555

45 9
0.9
50 10
  
 
 
 
|
P P D
P D P
P P


45/555
70/555

45 9
0.643
70 14
  
a. Interpretation: A subject who uses drugs has a
0.9 probability of getting a positive test result.
b. Interpretation: A
subject who gets a positive
test result, there is a 0.643
probability that this subject
actually uses drugs.
Note that P(positive test
result | subject uses drugs)
≠ P(subject uses drugs |
positive test result).

14. 14
Confusion of the Inverse: In general, P(B | A) ≠ P(A | B).
There could be individual cases where P(A | B) and P(B | A ) are equal, but they
are generally not equal.
Incorrect assumption of: P(B | A) = P(A | B) is called confusion of the inverse.
Example 7: Consider these events: D: It is dark outdoors. M: It is midnight.
Ignore the Alaskan winter and other such anomalies.
P(D | M) = 1 (It is certain to be dark given that it is midnight.)
P(M | D) = 0 (The probability that it is exactly midnight given that it
dark is almost zero.)
Here, P(D | M) ≠ P(M | D).
Confusion of the inverse occurs when we incorrectly switch those
probability values or think that they are equal.