1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370...
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Transcript of 1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370...
1
Probability-Independence and Fundamental Rules
Dr. Jerrell T. Stracener, SAE Fellow
EMIS 7370 STAT 5340
Probability and Statistics for Scientists and Engineers
Leadership in Engineering
Department of Engineering Management, Information and Systems
SMU BOBBY B. LYLESCHOOL OF ENGINEERING
EMIS - SYSTEMS ENGINEERING PROGRAM
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Two events A and B in S are independent if, and only if,
P(A B) = P(A)P(B)
Definition - Independence
3
• If A, B and C are independent events, in S, then
P(A B C) = P(A)P(B)P(C)
• If A1, …, An are independent events in S, then
n
1ii
n
1ii APAP
Rules of Probability
4
• Mutually Exclusive Events
If A and B are any two events in S, then
P(A B) = 0
• Complementary Events
If A' is the complement of A, then
P(A') = 1 - P(A)
Rules of Probability
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• Rule:
If A and B are any two events in S, then
P(A B) = P(A) + P(B) - P(A B)
• Rule:
If A and B are mutually exclusive, then
P(A B) = P(A) + P(B)
Rules of Probability - Addition Rules
6
• RuleFor any 3 events, A, B and C in S,
P(A B C) = P(A) + P(B) + P(C)
- P(A B) - P(A C)
- P(B C) + P(A B C)
• RuleIf A, B and C are mutually exclusive events in S, then
P(A B C) = P(A) + P(B) + P(C)
Rules of Probability
Example: Coin Tossing Game A game is played as follows:
a) A player tosses a coin two times in sequence. If at least one head occurs the player wins. What is the probability of winning?
a) The game is modified as follows:A player tosses a coin. If a head occurs, the player wins. Otherwise, the coin is tossed again. If a head occurs, the player wins. What is the probability of winning?
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Example
An biased coin (likelihood of a head is 0.75) is tossed three times in sequence.
What is the probability that 2 heads will occur?
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• If A1, A2, ..., An are mutually exclusive, then
P(A1 A2 ... An) = P(A1) + P(A2) + ... + P(An)
• Rule For events A1, A2, ... , An,
ji
ijji
n
1ii AAPAP
n
1ii
1n
kjiijk
kji AP1...AAAP
n
1iiAP
Rules of Probability continued
10
A town has two fire engines operating independently. The
probability that a specific fire engine is available when needed
is 0.99.
(a) What is the probability that neither is available when
needed?
(b) What is the probability that a fire engine is available
when needed?
Exercise - Fire Engines
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a. An electrical circuit consists of 4 switches in series. Assume that the operations of the 4 switches are statistically independent. If for each switch, the probability of failure (i.e., remaining open) is 0.02, what is the probability of circuit failure?
b. Rework for the case of 4 switches in parallel.
Exercise - 4 Switches in Series
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Series configuration
P(switch fails) = 0.02
P(switch does not fail) = 1 - 0.02 = 0.98
By observing the diagram above, the circuit fails if at least oneswitch fails. Also the circuit is a success if all 4 switches operatesuccessfully.
S1 S2 S3 S4
Exercise - 4 Switches in Series - solution
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Therefore,
P(circuit failure) = 1 - P(circuit success)
= 1 - P(S1 S2 S3 S4)
= 1 - (0.98)4
= 0.0776
Exercise - 4 Switches in Series - solution
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Parallel configuration
P(circuit failure) = P(4 of 4 switches fail)
= P(S1 S2 S3 S4)
= P(S1)P( S2)P( S3)P( S4)
= (0.02)4
= 0.00000016
S1
S2
S3
S4
Exercise - 4 Switches in Series - solution
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A rental car service facility has 10 foreign cars and 15 domesticcars waiting to be serviced on a particular Saturday morning. Because there are so few mechanics working on Saturday, only6 can be serviced. If the 6 are chosen at random, what is the probability that at least 3 of the cars selected are domestic?
Example - Car Rental
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Let A = exactly 3 of the 6 cars chosen are domestic. Assumingthat any particular set of 6 cars is as likely to be chosen as is anyother set of 6, we have equally likely outcomes, so
where n is the number of ways of choosing 6 cars from the 25and nA is the number of ways of choosing 3 domestic carsand 3 foreign cars. Thus
To obtain nA, think of first choosing 3 of the 15 domestic cars
6
25
n
nAP A
n
Example - Car Rental - solution
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and then 3 of the foreign cars. There are ways of choosing the3 domestic cars, and there are ways of choosing the 3foreign cars; nA is the product of these two numbers (visualize a tree diagram) using a product rule, so
!19!6
!25!7!3!10
!12!3!15
6
25
3
10
3
15
n
nAP A
3
15
3
10
3083.0
Example - Car Rental - solution
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Let D4 = (exactly 4 of the 6 cars chosen are domestic), and defineD5 and D6 in an analogous manner. Then the probability that atleast 3 domestic cars are selected is
P(D3 D4 D5 D6) = P(D3) + P(D4) + P(D5) + P(D6)
6
25
0
10
6
15
6
25
1
10
5
15
6
25
2
10
4
15
6
25
3
10
3
15
8530.0
This is also the probability that at most 3 foreign cars are selected.
Example - Car Rental - solution