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Introduction to Probability & Statistics South Dakota School of Mines & Technology Introduction to Probability & Statistics Industrial Engineering
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Concepts of Probability Introduction to Probability & Statistics Concepts of Probability
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Probability Concepts S = Sample Space : the set of all possible unique outcomes of a repeatable experiment. Ex: flip of a coin S = {H,T} No. dots on top face of a die S = {1, 2, 3, 4, 5, 6} Body Temperature of a live human S = [88,108]
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Probability Concepts Event : a subset of outcomes from a sample space. Simple Event: one outcome; e.g. get a 3 on one throw of a die A = {3} Composite Event: get 3 or more on throw of a die A = {3, 4, 5, 6}
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Rules of Events Union : event consisting of all outcomes present in one or more of events making up union. Ex: A = {1, 2}B = {2, 4, 6} A B = {1, 2, 4, 6}
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Rules of Events Intersection : event consisting of all outcomes present in each contributing event. Ex: A = {1, 2}B = {2, 4, 6} A B = {2}
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Rules of Events Complement : consists of the outcomes in the sample space which are not in stipulated event Ex: A = {1, 2}S = {1, 2, 3, 4, 5, 6} A = {3, 4, 5, 6}
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Rules of Events Mutually Exclusive : two events are mutually exclusive if their intersection is null Ex: A = {1, 2, 3}B = {4, 5, 6} A B = { } =
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Probability Defined u Equally Likely Events If m out of the n equally likely outcomes in an experiment pertain to event A, then p(A) = m/n
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Probability Defined u Equally Likely Events If m out of the n equally likely outcomes in an experiment pertain to event A, then p(A) = m/n Ex: Die example has 6 equally likely outcomes: p(2) = 1/6 p(even) = 3/6
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Probability Defined u Suppose we have a workforce which is comprised of 6 technical people and 4 in administrative support.
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Probability Defined u Suppose we have a workforce which is comprised of 6 technical people and 4 in administrative support. P(technical)= 6/10 P(admin) = 4/10
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Rules of Probability Let A = an event defined on the event space S 1.0 < P(A) < 1 2.P(S) = 1 3.P( ) = 0 4.P(A) + P( A ) = 1
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Addition Rule P(A B) = P(A) + P(B) - P(A B) AB
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Addition Rule P(A B) = P(A) AB
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Addition Rule P(A B) = P(A) + P(B) AB
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Addition Rule P(A B) = P(A) + P(B) - P(A B) AB
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Example u Suppose we have technical and administrative support people some of whom are male and some of whom are female.
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Example (cont) u If we select a worker at random, compute the following probabilities: P(technical) = 18/30
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Example (cont) u If we select a worker at random, compute the following probabilities: P(female) = 14/30
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Example (cont) u If we select a worker at random, compute the following probabilities: P(technical or female) = 22/30
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Example (cont) u If we select a worker at random, compute the following probabilities: P(technical and female) = 10/30
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u Alternatively we can find the probability of randomly selecting a technical person or a female by use of the addition rule. = 18/30 + 14/30 - 10/30 = 22/30 Example (cont) )()()()(FTPFPTPFTP -+=
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Operational Rules Mutually Exclusive Events: P(A B) = P(A) + P(B) AB
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Conditional Probability Suppose we look at the intersection of two events A and B. AB
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Conditional Probability Now suppose we know that event A has occurred. What is the probability of B given A? A A B P(B|A) = P( A B)/P(A)
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Example u Returning to our workers, suppose we know we have a technical person.
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Example u Returning to our workers, suppose we know we have a technical person. Then, P(Female | Technical) = 10/18
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Example u Alternatively, P(F | T) = P(F T) / P(T) = (10/30) / (18/30) = 10/18
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Independent Events u Two events are independent if P(A|B) = P(A) or P(B|A) = P(B) In words, the probability of A is in no way affected by the outcome of B or vice versa.
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Example u Suppose we flip a fair coin. The possible outcomes are HT The probability of getting a head is then P(H) = 1/2
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Example u If the first coin is a head, what is the probability of getting a head on the second toss? H,H H,T T,HT,T P(H 2 |H 1 ) = 1/2
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Example u If the first coin is a head, what is the probability of getting a head on the second toss? H,H H,T T,HT,T P(H 2 |H 1 ) = 1/2 = P(H 2 ) Tosses are independent
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Multiplication Rule P(B|A) = P( A B)/P(A) P(A B) = P(A)P(B|A)
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Multiplication Rule P(B|A) = P( A B)/P(A) P(A B) = P(A)P(B|A) Independence : P(B|A) = P(B) P(A B) = P(A)P(B)
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Example u Suppose we flip a fair coin twice. The possible outcomes are: H,H H,T T,HT,T P(2 heads) = P(H,H) = 1/4
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Example u Alternatively P(2 heads) = P(H 1 H 2 ) = P(H 1 )P(H 2 |H 1 ) = P(H 1 )P(H 2 ) = 1/2 x 1/2 = 1/4
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Example u Suppose we have a workforce consisting of male technical people, female technical people, male administrative support, and female administrative support. Suppose the make up is as follows Tech Admin Male Female 8 10 8 4
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Example Let M = male, F = female, T = technical, and A = administrative. Compute the following: P(M T) = ? P(T|F) = ? P(M|T) = ? Tech Admin Male Female 8 10 8 4
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Example Let M = male, F = female, T = technical, and A = administrative. Compute the following: P(M T) = 8/30 P(T|F) = 10/14 P(M|T) = 8/18 Tech Admin Male Female 8 10 8 4
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Introduction to Probability & Statistics South Dakota School of Mines & Technology Introduction to Probability & Statistics Industrial Engineering
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Counting Introduction to Probability & Statistics Counting
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Fundamental Rule u If an action can be performed in m ways and another action can be performed in n ways, then both actions can be performed in m n ways.
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Fundamental Rule u Ex: A lottery game selects 3 numbers between 1 and 5 where numbers can not be selected more than once. If the game is truly random and order is not important, how many possible combinations of lottery numbers are there?
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Fundamental Rule u Ex: A lottery game selects 3 numbers between 1 and 5 where numbers can not be selected more than once. If the game is truly random and order is not important, how many possible combinations of lottery numbers are there? 1234512345
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Fundamental Rule u Ex: A lottery game selects 3 numbers between 1 and 5 where numbers can not be selected more than once. If the game is truly random and order is not important, how many possible combinations of lottery numbers are there? 1234512345 23452345
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Fundamental Rule u Ex: A lottery game selects 3 numbers between 1 and 5 where numbers can not be selected more than once. If the game is truly random and order is not important, how many possible combinations of lottery numbers are there? 1234512345 23452345 345345
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Fundamental Rule u Ex: A lottery game selects 3 numbers between 1 and 5 where numbers can not be selected more than once. If the game is truly random and order is not important, how many possible combinations of lottery numbers are there? 1234512345 23452345 345345 LN = 543 = 60
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Combinations u Suppose we flip a coin 3 times, how many ways are there to get 2 heads?
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Combinations u Suppose we flip a coin 3 times, how many ways are there to get 2 heads? Soln: List all possibilities: H,H,HH,T,T H,H,TH,T,H H,T,HT,H,H T,H,HT,T,T
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Combinations Of 8 possible outcomes, 3 meet criteria H,H,HH,T,T H,H,TH,T,H H,T,HT,H,H T,H,HT,T,T
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Combinations If we don’t care in which order these 3 occur H,H,T H,T,H T,H,H Then we can count by combination.
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Combinations u Combinations n C k = the number of ways to count k items out n total items order not important. n = total number of items k = number of items pertaining to event A
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Example u How many ways can we select a 4 person committee from 10 students available?
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Example u How many ways can we select a 4 person committee from 10 students available? No. Possible Committees =
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Example u We have 20 students, 8 of whom are female and 12 of whom are male. How many committees of 5 students can be formed if we require 2 female and 3 male?
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Example u We have 20 students, 8 of whom are female and 12 of whom are male. How many committees of 5 students can be formed if we require 2 female and 3 male? Soln: Compute how many 2 member female committees we can have and how many 3 member male committees. Each female committee can be combined with each male committee.
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Example
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Permutations u Permutations is somewhat like combinations except that order is important.
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Example u How many ways can a four member committee be formed from 10 students if the first is President, second selected is Vice President, 3rd is secretary and 4th is treasurer?
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Example u How many ways can a four member committee be formed from 10 students if the first is President, second selected is Vice President, 3rd is secretary and 4th is treasurer?
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Example u How many ways can a four member committee be formed from 10 students if the first is President, second selected is Vice President, 3rd is secretary and 4th is treasurer? 10 P 4 = 10*9*8*7 = 5,040
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Introduction to Probability & Statistics South Dakota School of Mines & Technology Introduction to Probability & Statistics Industrial Engineering
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Random Variables Introduction to Probability & Statistics Random Variables
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Random Variables A Random Variable is a function that associates a real number with each element in a sample space. Ex: Toss of a die X = # dots on top face of die = 1, 2, 3, 4, 5, 6
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Random Variables A Random Variable is a function that associates a real number with each element in a sample space. Ex: Flip of a coin 0, heads X = 1, tails
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Random Variables A Random Variable is a function that associates a real number with each element in a sample space. Ex: Flip 3 coins 0 if TTT X = 1 if HTT, THT, TTH 2 if HHT, HTH, THH 3 if HHH
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Random Variables A Random Variable is a function that associates a real number with each element in a sample space. Ex: X = lifetime of a light bulb X = [0, )
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Distributions Let X= number of dots on top face of a die when thrown p(x) = Prob{X=x} x 1 2 3 4 5 6 p(x) 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6
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Cumulative Let F(x) = Pr{X < x} x 1 2 3 4 5 6 p(x) 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6 F(x) 1 / 6 2 / 6 3 / 6 4 / 6 5 / 6 6 / 6
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Complementary Cumulative Let F(x) = 1 - F(x) = Pr{X > x} x 1 2 3 4 5 6 p(x) 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6 F(x) 1 / 6 2 / 6 3 / 6 4 / 6 5 / 6 6 / 6 F(x) 5 / 6 4 / 6 3 / 6 2 / 6 1 / 6 0 / 6
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Discrete Univariate u Binomial u Discrete Uniform (Die) u Hypergeometric u Poisson u Bernoulli u Geometric u Negative Binomial
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Binomial u What is the probability of getting 2 heads out of 3 flips of a coin?
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Binomial u What is the probability of getting 2 heads out of 3 flips of a coin? Soln: H,H,HH,T,T H,H,TT,H,T H,T,HT,T,H T,H,HT,T,T
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Binomial P{2 heads in 3 flips} = P{H,H,T} + P{H,T,H} + P{T,H,H} = 3 P{H}P{H}P{T} = 3 C 2 P{H} 2 P{T} 3-2 = 3 C 2 p 2 (1-p) 3-2
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Distributions Binomial : X = number of successes in n bernoulli trials p = Pr(success) = const. from trial to trial n = number of trials p(x) = b(x; n,p) = n xnx pp xnx ! !()! () 1
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Binomial Distribution 0.0 0.1 0.2 0.3 0.4 0.5 012345 x P(x) 0.0 0.1 0.2 0.3 0.4 0.5 012345 x P(x) n=5, p=.3 n=8, p=.5 x 0.0 0.1 0.2 0.3 0.4 0.5 012345678 P(x) n=4, p=.8 0.0 0.1 0.2 0.3 0.4 0.5 024 x P(x) n=20, p=.5
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Example u Suppose we manufacture circuit boards with 95% reliability. If approximately 5 circuit boards in 100 are defective, what is the probability that a lot of 10 circuit boards has one or more defects?
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Example (soln.) Pr{} }XX 110 1 10 0 0595) 010 ! !(!) (.) = 1 -.95 10 =.4013
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Example For p nPr{X > 1}.05 10 0.4013.05 100 0.9941.051,000 1.0000.01 10 0.0956.01 100 0.6340.011,000 1.0000
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99% Defect Free Rate u 500 incorrect surgical procedures every week u 20,000 prescriptions filled incorrectly each year u 12 babies given to the wrong parents each day u 16,000 pieces of mail lost each hour u 2 million documents lost by IRS each year u 22,000 checks deducted from wrong accounts during next hour (Ref: Quality, March 91)
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Example Educational curriculum is typically aimed at the normed population which tends to be convergent in nature. Educational research also tells us this curriculum is inappropriate for a student two standard deviations above the norm since these students tend to be more divergent in their thinking. If approximately 3% of the population lies 2 standard deviations above the norm, what is the probability that a school with student population of 100 students has one or more “gifted” students.
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Example Pr{} }XX 110 1 100 0 0397) 0100 ! !(!) (.) = 1 -.97 100 =.9524
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Binomial Measures Mean : Variance: xpx x () 22 ()()xpx x = np = np(1-p)
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Binomial Distribution 0.0 0.1 0.2 0.3 0.4 0.5 012345 x P(x) 0.0 0.1 0.2 0.3 0.4 0.5 012345 x P(x) n=5, p=.3 n=8, p=.5 x 0.0 0.1 0.2 0.3 0.4 0.5 012345678 P(x) n=4, p=.8 0.0 0.1 0.2 0.3 0.4 0.5 024 x P(x) n=20, p=.5
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Continuous Distribution x a b c d f(x) A 1.f(x) > 0, all x 2. 3.P(A) = Pr{a < x < b} = 4.Pr{X=a} = fxdx a d () 1 fx b c () fx a a () 0
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Continuous Univariate u Normal u Uniform u Exponential u Weibull u LogNormal u Beta u T-distribution u Chi-square u F-distribution u Maxwell u Raleigh u Triangular u Generalized Gamma u H-function
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Normal Distribution 65% 95% 99.7%
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Scale Parameter > 1 = 1
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Location Parameter x x > 1 = 1
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Std. Normal Transformation Standard Normal Z X f(z) N(0,1)
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Example u Suppose a resistor has specifications of 100 + 10 ohms. R = actual resistance of a resistor and R N(100,5). What is the probability a resistor taken at random is out of spec? x LSLUSL 100
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Example Cont. x LSLUSL 100 Pr{in spec}= Pr{90 < x < 110} Pr 90100 5 110100 5 x = Pr(-2 < z < 2)
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Example Cont. x LSLUSL 100 Pr{in spec} = Pr(-2 < z < 2) = [F(2) - F(-2)] = (.9773 -.0228) =.9545
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Example Cont. x LSLUSL 100 Pr{in spec} = Pr(-2 < z < 2) = [F(2) - F(-2)] = (.9773 -.0228) =.9545 Pr{out of spec} = 1 - Pr{in spec} = 1 -.9545 = 0.0455
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Example u Assume that the per capita income in South Dakota is normally distributed with a mean of $20,000 and a standard deviation of $4,000. If the poverty level is considered to be $15,000 per year, compute the percentage of South Dakotans who would be considered to be at or below the poverty level.
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Example Pr{poverty level} = Pr{X < 15,000} = Pr{Z < -1.25} = 0.5 - Pr{0 < Z < 1.25} = 0.5 - 0.3944 = 0.1056 x 15,000 20,000
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Other Continuous Distributions
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Exponential Distribution fxe x () Density Cumulative Mean 1/ Variance 1/ 2 Fxe x () 1, x > 0 0.0 0.5 1.0 00.511.522.53 Time to Fail Density =1
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Exponential Distribution fxe x () Density Cumulative Mean 1/ Variance 1/ 2 Fxe x () 1, x > 0 =1 0.0 0.5 1.0 1.5 2.0 00.511.522.53 Time to Fail Density =2
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0.0 0.5 1.0 0.01.02.03.04.05.06.07.0 x f(x) LogNormal Density Cumulative no closed form Mean Variance fx x e x () ln 1 2 1 2 2 , x > 0 e 2 2 ee 2 22 1 () = 0 =1
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0.0 0.5 1.0 0.01.02.03.04.05.06.07.0 x f(x) LogNormal Density Cumulative no closed form Mean Variance fx x e x () ln 1 2 1 2 2 , x > 0 e 2 2 ee 2 22 1 () = 0 =2
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LogNormal Density Cumulative no closed form Mean Variance fx x e x () ln 1 2 1 2 2 , x > 0 e 2 2 ee 2 22 1 () = 0 =0.5
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Gamma Density Cumulative no closed form for integer Mean Variance 2 fxxe x () () / 1, x > 0 0.0 0.5 1.0 0.01.02.03.04.05.06.07.0 x f(x) =1
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0.0 0.5 1.0 0.01.02.03.04.05.06.07.0 x f(x) Gamma Density Cumulative no closed form for integer Mean Variance 2 fxxe x () () / 1, x > 0 =2
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0.0 0.5 1.0 0.01.02.03.04.05.06.07.0 x f(x) Gamma Density Cumulative no closed form for integer Mean Variance 2 fxxe x () () / 1, x > 0 =3
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0.0 0.5 1.0 1.5 0.01.02.03.04.05.06.07.0 Weibull Density Cumulative Mean Variance fxxe x () (/) 21 2, x > 0 Fxe x () (/) 1 2 1 2 2 2 211 = 1 = 1
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0.0 0.5 1.0 1.5 0.01.02.03.04.05.06.07.0 Weibull Density Cumulative Mean Variance fxxe x () (/) 21 2, x > 0 Fxe x () (/) 1 2 1 2 2 2 211 = 1 = 2
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0.0 0.5 1.0 1.5 0.01.02.03.04.05.06.07.0 Weibull Density Cumulative Mean Variance fxxe x () (/) 21 2, x > 0 Fxe x () (/) 1 2 1 2 2 2 211 = 1 = 3
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Uniform Density Cumulative Mean (a + b)/2 Variance (b - a) 2 /12 fx ba () 1, a < x < b Fx xa ba () f(x) x a b
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End Probability Review Session 1
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