Additive Rule/ Contingency Table
Experiment: Draw 1 card from a standard 52 card deck. Record
Value (A-K), Color & Suit.
 The probabilities associated with drawing an ace and with
drawing a black card are shown in the following contingency
table:
Color
Type
Ace


Red
Black
Total
2
2
4
Non-Ace
24
24
48
Total
26
26
52
Event A = ace Event B = black card
Therefore the probability of drawing an ace or a black card is:
4 26 2
28
P ( A  B )  P ( A )  P (B )  P ( A  B ) 



52 52 52 52
JMB Ch2 Lecture 2
EGR 252.003 Fall2016
Slide 1
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 defect 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 (HJ)
0.46
Oven/MW junction (OM)
0.14
Thermostat (T)
0.09
Oven coil (OC)
0.24
Electronic controls (EC)
0.07
 The sum of the probabilities equals _____
JMB Ch2 Lecture 2
EGR 252.003 Fall2016
Slide 2
Short Circuit Example - Probabilities
 If we are told that the probabilities represent
mutually exclusive events, we can calculate the
following:
 The probability that the short circuit does not occur
at the house junction is
P(HJ’) = 1 - P(HJ) = 1 – 0.46 = 0.54
 The probability that the short circuit occurs at either
the Oven/MW junction or the oven coil is
P(OM U OC) = P(OM)+P(OC) = 0.14 + 0.24 = 0.38
JMB Ch2 Lecture 2
EGR 252.003 Fall2016
Slide 3
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
 Event B = odd number
 (B∩A) = {7, 9, 11}
P(A) = 4/10
P(B) = 6/10
P (B∩A) = 3/10
 P(B|A) = P(B ∩ A) / P(A) = (3/10) / (4/10) = 3/4
JMB Ch2 Lecture 2
EGR 252.003 Fall2016
Slide 4
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
JMB Ch2 Lecture 2
EGR 252.003 Fall2016
Slide 5
Independence Definitions
If 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)
JMB Ch2 Lecture 2
EGR 252.003 Fall2016
Slide 6
Independence Example


A quality engineer collected the following data on 100 defective
items produced by a manufacturer in the southeast:
Mechanical
Other
Day
20
15
25
Night
10
20
10
P(Day) = (20+15+25) / 100 = .60 or 60%
What was the relative frequency of defectives categorized as
electrical?


Electrical
What is the probability that the defective items were associated with
the day shift?


Problem/Shift
(20 + 10) / 100
P(Electrical) = .30
Are Electrical and Day independent?


P(E ∩ D) = 20 / 100 = .20
P(D) P(E) = (.60) (.30) = .18
Since .20 ≠.18, Day and Electrical are not independent.
JMB Ch2 Lecture 2
EGR 252.003 Fall2016
Slide 7
Serial and Parallel Systems
 For increased safety and reliability, systems are often designed with
redundancies. A typical system might look like the following:
0.88
C
0.95
0.9
0.97
A
B
E
0.85
D
 Principles:
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 Ch2 Lecture 2
EGR 252.003 Fall2016
Slide 8
Serial and Parallel Systems
0.88
C
0.95
0.9
A
B
0.97
E
0.85
D
 What is the probability that:
 Segment 1 works?
 A and B in series
P(A∩B) = P(A) * P(B) = (0.95)(0.90) = 0.855
1
2
 Segment 2 works?
 C and D in parallel will work unless both C and D do not function
1 – P(C’) * P(D’) = 1 – (0.12) * (0.15) = 1 - 0.018 = 0.982
 The entire system works?
 Segment 1, Segment 2 and E in series
P(Segment1) * P(Segment2) * P(E) = 0.855*0.982*0.97 = 0.814
JMB Ch2 Lecture 2
EGR 252.003 Fall2016
Slide 9