1 probability- independence and fundamental rules dr. jerrell t. stracener, sae fellow emis 7370...
TRANSCRIPT
![Page 1: 1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370 STAT 5340 Probability and Statistics for Scientists and](https://reader031.vdocuments.site/reader031/viewer/2022032104/56649c755503460f94929248/html5/thumbnails/1.jpg)
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
![Page 2: 1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370 STAT 5340 Probability and Statistics for Scientists and](https://reader031.vdocuments.site/reader031/viewer/2022032104/56649c755503460f94929248/html5/thumbnails/2.jpg)
2
Two events A and B in S are independent if, and only if,
P(A B) = P(A)P(B)
Definition - Independence
![Page 3: 1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370 STAT 5340 Probability and Statistics for Scientists and](https://reader031.vdocuments.site/reader031/viewer/2022032104/56649c755503460f94929248/html5/thumbnails/3.jpg)
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
![Page 4: 1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370 STAT 5340 Probability and Statistics for Scientists and](https://reader031.vdocuments.site/reader031/viewer/2022032104/56649c755503460f94929248/html5/thumbnails/4.jpg)
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
![Page 5: 1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370 STAT 5340 Probability and Statistics for Scientists and](https://reader031.vdocuments.site/reader031/viewer/2022032104/56649c755503460f94929248/html5/thumbnails/5.jpg)
5
• 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
![Page 6: 1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370 STAT 5340 Probability and Statistics for Scientists and](https://reader031.vdocuments.site/reader031/viewer/2022032104/56649c755503460f94929248/html5/thumbnails/6.jpg)
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
![Page 7: 1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370 STAT 5340 Probability and Statistics for Scientists and](https://reader031.vdocuments.site/reader031/viewer/2022032104/56649c755503460f94929248/html5/thumbnails/7.jpg)
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?
![Page 8: 1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370 STAT 5340 Probability and Statistics for Scientists and](https://reader031.vdocuments.site/reader031/viewer/2022032104/56649c755503460f94929248/html5/thumbnails/8.jpg)
8
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?
![Page 9: 1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370 STAT 5340 Probability and Statistics for Scientists and](https://reader031.vdocuments.site/reader031/viewer/2022032104/56649c755503460f94929248/html5/thumbnails/9.jpg)
9
• 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
![Page 10: 1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370 STAT 5340 Probability and Statistics for Scientists and](https://reader031.vdocuments.site/reader031/viewer/2022032104/56649c755503460f94929248/html5/thumbnails/10.jpg)
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
![Page 11: 1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370 STAT 5340 Probability and Statistics for Scientists and](https://reader031.vdocuments.site/reader031/viewer/2022032104/56649c755503460f94929248/html5/thumbnails/11.jpg)
11
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
![Page 12: 1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370 STAT 5340 Probability and Statistics for Scientists and](https://reader031.vdocuments.site/reader031/viewer/2022032104/56649c755503460f94929248/html5/thumbnails/12.jpg)
12
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
![Page 13: 1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370 STAT 5340 Probability and Statistics for Scientists and](https://reader031.vdocuments.site/reader031/viewer/2022032104/56649c755503460f94929248/html5/thumbnails/13.jpg)
13
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
![Page 14: 1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370 STAT 5340 Probability and Statistics for Scientists and](https://reader031.vdocuments.site/reader031/viewer/2022032104/56649c755503460f94929248/html5/thumbnails/14.jpg)
14
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
![Page 15: 1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370 STAT 5340 Probability and Statistics for Scientists and](https://reader031.vdocuments.site/reader031/viewer/2022032104/56649c755503460f94929248/html5/thumbnails/15.jpg)
15
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
![Page 16: 1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370 STAT 5340 Probability and Statistics for Scientists and](https://reader031.vdocuments.site/reader031/viewer/2022032104/56649c755503460f94929248/html5/thumbnails/16.jpg)
16
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
![Page 17: 1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370 STAT 5340 Probability and Statistics for Scientists and](https://reader031.vdocuments.site/reader031/viewer/2022032104/56649c755503460f94929248/html5/thumbnails/17.jpg)
17
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
![Page 18: 1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370 STAT 5340 Probability and Statistics for Scientists and](https://reader031.vdocuments.site/reader031/viewer/2022032104/56649c755503460f94929248/html5/thumbnails/18.jpg)
18
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