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Additive Rule/ Contingency Table

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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 ... – PowerPoint PPT presentation

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Title: Additive Rule/ Contingency Table


1
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
  • Event A ace Event B black card
  • Therefore the probability of drawing an ace or a
    black card is

Type Color Color Total
Type Red Black Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
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 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
3
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

4
Conditional Probability
  • The conditional probability of B given A is
    denoted by P(BA) and is calculated by
  • P(BA) P(B n 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
  • (BnA) 7, 9, 11 P (BnA) 3/10
  • P(BA) P(B n A) / P(A) (3/10) / (4/10) 3/4

5
Multiplicative Rule
  • If in an experiment the events A and B can both
    occur, then
  • P(B n A) P(A) P(BA)
  • 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(BA) 3/4 (calculated in previous slide)
  • P(BnA) P(A)P(BA) (4/10)(3/4) 3/10

6
Independence Definitions
  • If in an experiment the conditional probabilities
    P(AB) and P(BA) exist, the events A and B are
    independent if and only if
  • P(AB) P(A) or P(BA) P(B)
  • Two events A and B are independent if and only if
    P A n B P(A) P(B)

7
Independence Example
  • A quality engineer collected the following data
    on 100 defective items produced by a manufacturer
    in the southeast

Problem/Shift Electrical Mechanical Other
Day 20 15 25
Night 10 20 10
  • What is the probability that the defective items
    were associated with the day shift?
  • P(Day) (201525) / 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 n D) 20 / 100 .20 P(D) P(E)
    (.60) (.30) .18
  • Since .20 ?.18, Day and Electrical are not
    independent.

8
Serial and Parallel Systems
  • For increased safety and reliability, systems are
    often designed with redundancies. A typical
    system might look like the following

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.
9
Serial and Parallel Systems
1
  • What is the probability that
  • Segment 1 works?
  • A and B in series
  • P(AnB) 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)
    0.8550.9820.97 0.814

2
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