Additive Rule/ Contingency Table

JMB Ch2 Lecture 2 vSp2012 EGR 252.001 2012 Slide 1

Additive Rule/ Contingency TableExperiment: 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:

Event A = ace Event B = black card Therefore the probability of drawing an ace or a black card is:

Type

Color

TotalRed Black

Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52

52

28

52

2

52

26

52

4)()()()( BAPBPAPBAP


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:

The sum of the probabilities equals _____

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


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


Conditional ProbabilityThe 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) = 4/10Event 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


Multiplicative RuleIf 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) = 4/10Event B = odd number P(B) = 6/10P(B|A) = 3/4 (calculated in previous slide)

P(B∩A) = P(A)*P(B|A) = (4/10)*(3/4) = 3/10


Independence Definitions

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)


Independence Example A quality engineer collected the following data on 100 defective

items produced by a manufacturer in the southeast:

What is the probability that the defective items were associated with the day shift? P(Day) = (20+15+25) / 100 = .60 or 60%

What was the relative frequency of defectives categorized as electrical? (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.

Problem/Shift Electrical Mechanical Other

Day 20 15 25

Night 10 20 10


Serial and Parallel Systems For increased safety and reliability, systems are often designed with

redundancies. A typical system might look like the following:

Principles:

0.95 0.9

0.88

0.85

0.97

A B

C

D

E

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.


Serial and Parallel Systems

What is the probability that: Segment 1 works? A and B in series

P(A∩B) = P(A) * P(B) = (0.95)(0.9) = 0.855

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

0.95 0.9

0.88

0.85

0.97

A B

C

D

E

12

Additive Rule/ Contingency Table

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