1. Additive Rule Review Experiment: Draw 1 Card. Note Kind, Color & Suit.
Probabilities associated with drawing an ace and with drawing a black card are shown in the following contingency table:
Therefore the probability of drawing an ace or a black card is given by:
Type
Color
Total
Red
Black
Ace 2 4
Non-Ace 24 48 26 52
JMB Chapter 2 Lecture 2
EGR Spring 2008

2. Short Circuit Example - Data
An appliance manufacturer has learned of an increased incidence of short circuits and fires in a line of ranges sold over a 5 month period. A review of the FMEA data indicates the probabilities that if a short circuit occurs, it will be at any one of several locations is as follows:
Location P
House Junction
0.46
Oven/MW junction
0.14
Thermostat
0.09
Oven coil
0.24
Electronic controls
0.07
JMB Chapter 2 Lecture 2
EGR Spring 2008

3. Short Circuit Example - Probabilities
The probability that the short circuit does not occur at the house junction is …
P(H’) = 1 – 0.46 = 0.54
The probability that the short circuit occurs at either the Oven/MW junction or the oven coil is …
P(J U C) = P(J) + P(C) = = 0.38
P(H’) = 1 – 0.46 = 0.54
P(J U C) = P(J) + P(C) = = 0.38
P(E ∩ T) = P(E)*P(T) = * 0.07 =
JMB Chapter 2 Lecture 2
EGR Spring 2008

4. Conditional Probability
The conditional probability of B given A is denoted by P(B|A) and is calculated by
P(B|A) = P(B ∩ A) / P(A)
Example:
S = {1,2,3,4,5,6,7,8,9,11}
Event A = number greater than 6 P(A) = 4/10
Event B = odd number P(B) = 6/10
(B∩A) = {7, 9, 11} P (B∩A) = 3/10
P(B|A) = P(B ∩ A) / P(A) = (3/10) / (4/10) = 3/4
P(C’|2) = P(C’∩D)/P(2) = ((1-.88)*0.85)/0.982 =
In general, the conditional probability of B given A is denoted by P(B|A) and is given by
P(B|A) = P(B ∩ A) / P(A)
(Definition 2.9, pg. 48)
JMB Chapter 2 Lecture 2
EGR Spring 2008

5. Multiplicative Rule
If in an experiment the events A and B can both occur, then
P(B ∩ A) = P(A) * P(B|A)
Previous Example:
S = {1,2,3,4,5,6,7,8,9,11}
Event A = number greater than 6 P(A) = 4/10
Event B = odd number P(B) = 6/10
P(B|A) = 3/4 (calculated in previous slide)
P(B∩A) = P(A)*P(B|A) = (4/10)*(3/4) = 3/10
P(C’|2) = P(C’∩D)/P(2) = ((1-.88)*0.85)/0.982 =
In general, the conditional probability of B given A is denoted by P(B|A) and is given by
P(B|A) = P(B ∩ A) / P(A)
(Definition 2.9, pg. 48)
JMB Chapter 2 Lecture 2
EGR Spring 2008

6. Independence
If in an experiment the conditional probabilities P(A|B) and P(B|A) exist, the events A and B are independent if and only if
P(A|B) = P(A) or P(B|A) = P(B)
Two events A and B are independent if and only if P A ∩ B = P(A) P(B)
P(C’|2) = P(C’∩D)/P(2) = ((1-.88)*0.85)/0.982 =
In general, the conditional probability of B given A is denoted by P(B|A) and is given by
P(B|A) = P(B ∩ A) / P(A)
(Definition 2.9, pg. 48)
JMB Chapter 2 Lecture 2
EGR Spring 2008

7. Independence Example
A quality engineer collected the following data on defects:
Electrical
Mechanical
Other
Day 20 15 25
Night 10
What is the likelihood that defects were associated with the day shift?
P(Day) = ( ) / 100 = .60
What was the relative frequency of electrical defects?
( ) / P(Electrical) = .30
Are Electrical and Day independent?
P(E ∩ D) = 20 / 100 = P(D) P(E) = (.60) (.30) = .18
Since .20 ≠.18, Day and Electrical are not independent.
JMB Chapter 2 Lecture 2
EGR Spring 2008

8. Serial and Parallel Systems
For increased safety and reliability, systems are often designed with redundancies. A typical system might look like the following:
NOTE: if components are in serial (e.g., A & B), all must work in order for the system to work. If components are in parallel, the system works if any of the components work.
If components are in serial (e.g., A & B), all must work in order for the system to work.
If components are in parallel, the system works if any of the components work.
JMB Chapter 2 Lecture 2
EGR Spring 2008

9. Serial and Parallel Systems1
What is the probability that:
Segment 1 works?
P(A∩B) = P(A)P(B) = (.95)(.9) = .855
Segment 2 works?
1 – P(C’)P(D’) = 1 – (.12)(.15) = = 0.982
The entire system works?
P(Segment1)P(Segment2)P(E) = .855*.982*.97 =.814 2
Segment 1: P(A∩B) = P(A)P(B) = (.95)(.9) = .855
Segment 2: 1 – P(C’)P(D’) = 1 – (.12)(.15) = = 0.982
Entire system: P(1)P(2)P(E) = .855*.982*.97 =.814
JMB Chapter 2 Lecture 2
EGR Spring 2008