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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

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• 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

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• 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

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• 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

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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