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## Introduction to Probability & Statistics Concepts of Probability

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**South Dakota School of Mines & TechnologyIntroduction to**Probability & StatisticsIndustrial Engineering**Introduction to Probability & StatisticsConcepts of**Probability**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]**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}**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}**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}**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}**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 = { } = **Probability Defined**• Equally Likely Events If m out of the n equally likely outcomes in an experiment pertain to event A, then p(A) = m/n**Probability Defined**• 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**Probability Defined**• Suppose we have a workforce which is comprised of 6 technical people and 4 in administrative support.**Probability Defined**• 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**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**Addition Rule**P(A B) = P(A) + P(B) - P(A B) A B**Addition Rule**P(A B) = P(A) + P(B) - P(A B) A B**Example**• Suppose we have technical and administrative support people some of whom are male and some of whom are female.**Example (cont)**• If we select a worker at random, compute the following probabilities: P(technical) = 18/30**Example (cont)**• If we select a worker at random, compute the following probabilities: P(female) = 14/30**Example (cont)**• If we select a worker at random, compute the following probabilities: P(technical or female) = 22/30**Example (cont)**• If we select a worker at random, compute the following probabilities: P(technical and female) = 10/30**È**= + - Ç P ( T F ) P ( T ) P ( F ) P ( T F ) Example (cont) • 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**Operational Rules**Mutually Exclusive Events: P(A B) = P(A) + P(B) A B**A**Conditional Probability Now suppose we know that event A has occurred. What is the probability of B given A? A B P(B|A) = P(A B)/P(A)**Example**• Returning to our workers, suppose we know we have a technical person.**Example**• Returning to our workers, suppose we know we have a technical person. Then, P(Female | Technical) = 10/18**Example**• Alternatively, P(F | T) = P(F T) / P(T) = (10/30) / (18/30) = 10/18 Ç**Independent Events**• 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.**Example**• Suppose we flip a fair coin. The possible outcomes are H T The probability of getting a head is then P(H) = 1/2**Example**• If the first coin is a head, what is the probability of getting a head on the second toss? H,H H,T T,H T,T P(H2|H1) = 1/2**Example**• Suppose we flip a fair coin twice. The possible outcomes are: H,H H,T T,H T,T P(2 heads) = P(H,H) = 1/4**Example**• Alternatively P(2 heads) = P(H1 H2) = P(H1)P(H2|H1) = P(H1)P(H2) = 1/2 x 1/2 = 1/4**Tech Admin**Male 8 8 Female 10 4 Example • 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 8 8 Female 10 4 Example Let M = male, F = female, T = technical, and A = administrative. Compute the following: P(M T) = ? P(T|F) = ? P(M|T) = ?**South Dakota School of Mines & TechnologyIntroduction to**Probability & StatisticsIndustrial Engineering**Fundamental Rule**• 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.**Fundamental Rule**• 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?**Fundamental Rule**• 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? 1 2 3 4 5**Fundamental Rule**• 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? 2 3 4 5 1 2 3 4 5**3**4 5 2 3 4 5 1 2 3 4 5 Fundamental Rule • 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?**3**4 5 2 3 4 5 1 2 3 4 5 Fundamental Rule • 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? LN = 5•4•3 = 60**Combinations**• Suppose we flip a coin 3 times, how many ways are there to get 2 heads?**Combinations**• Suppose we flip a coin 3 times, how many ways are there to get 2 heads? Soln: List all possibilities: H,H,H H,T,T H,H,T H,T,H H,T,H T,H,H T,H,H T,T,T**Combinations**Of 8 possible outcomes, 3 meet criteria H,H,H H,T,T H,H,T H,T,H H,T,H T,H,H T,H,H T,T,T**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.**Combinations**• Combinations nCk = 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**Example**• How many ways can we select a 4 person committee from 10 students available?**Example**• How many ways can we select a 4 person committee from 10 students available? No. Possible Committees =**Example**• 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?