Download Probability - Independence & Fundamental Rules

Systems Engineering Program
Department of Engineering Management, Information and Systems
EMIS 7370/5370 STAT 5340 :
PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS
ProbabilityIndependence and Fundamental Rules
Dr. Jerrell T. Stracener, SAE Fellow
Leadership in Engineering
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Definition - Independence
Two events A and B in S are independent if, and only if,
P(A  B) = P(A)P(B)
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Rules of Probability
• 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
 n
P  A i    PA i 
 i 1  i 1
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Rules of Probability
• 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)
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Rules of Probability - Addition Rules
• 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)
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Rules of Probability
• Rule
For 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)
• Rule
If A, B and C are mutually exclusive events in S, then
P(A  B  C) = P(A) + P(B) + P(C)
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Example: Coin Tossing Game
A game is played as follows:
a) A player tosses a coin two times in sequence.
If at least on head occurs the player wins.
What is the probability of winning?
b) 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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Rules of Probability continued
• Rule
For events A1, A2, ... , An,
 n

P  A i  
 i 1 
 PA    PA
n
i 1
i
i
Aj
ij
i j
n


n 1
  PA i  A j  A k   ...   1 P  A i 
ijk
 i 1 
i  j k
• If A1, A2, ..., An are mutually exclusive, then
P(A1  A2  ...  An) = P(A1) + P(A2) + ... + P(An)
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Exercise - Fire Engines
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?
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Exercise - 4 Switches in Series
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.
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Exercise - 4 Switches in Series - solution
Series configuration
S1
S2
S3
S4
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
one
switch fails. Also the circuit is a success if all 4 switches
operate
successfully.
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Exercise - 4 Switches in Series - solution
Therefore,
P(circuit failure) = 1 - P(circuit success)
= 1 - P(S1  S2  S3  S4)
= 1 - (0.98)4
= 0.0776
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Exercise - 4 Switches in Series - solution
Parallel configuration
S1
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
S2
S3
S4
= 0.00000016
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Example - Car Rental
A rental car service facility has 10 foreign cars and 15 domestic
cars waiting to be serviced on a particular Saturday morning.
Because there are so few mechanics working on Saturday, only
6 can be serviced. If the 6 are chosen at random, what is the
probability that at least 3 of the cars selected are domestic?
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Example - Car Rental - solution
Let A = exactly 3 of the 6 cars chosen are domestic. Assuming
that any particular set of 6 cars is as likely to be chosen as is any
other set of 6, we have equally likely outcomes, so
nA
PA  
n
where n is the number of ways of choosing 6 cars from the 25
and nA is the number of ways of choosing 3 domestic cars
and 3 foreign cars. Thus
 25 
n 
 6 
 
To obtain nA, think of first choosing 3 of the 15 domestic cars
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Example - Car Rental - solution
15 
 
3
and then 3 of the foreign cars. There are ways of choosing the
10 
3 domestic cars, and there are  3  ways of choosing the 3
foreign cars; nA is the product of these two numbers
(visualize a tree diagram) using a product rule, so
15 10  15! 10!
  

n A  3  3  3!12! 3!7!
PA  


25!
n
 25 
 
6!19!
6
 0.3083
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Example - Car Rental - solution
Let D4 = (exactly 4 of the 6 cars chosen are domestic), and define
D5 and D6 in an analogous manner. Then the probability that at
least 3 domestic cars are selected is
P(D3  D4  D5  D6) = P(D3) + P(D4) + P(D5) + P(D6)
15 10  15 10  15 10  15 10 
           
3  3   4  2   5  1   6  0 





 25 
 25 
 25 
 25 
 
 
 
 
6
6
6
6
 0.8530
This is also the probability that at most 3 foreign cars are selected.
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