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Chapter 6 Probability Mohamed Elhusseiny mohuss@hotmail.com
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Introduction This chapter introduced the basic concepts of probability. It outlined rules and techniques for assigning probabilities to events. One objective in his and the following chapters is to develop the probability-based tools that are at the basis of statistical inference. Probability can also play a critical role in decision making
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Assigning Probabilities to Events Random Experiment A random experiment is an action or process that lead to one of several possible outcomes. –O–O–O–Or is an experiment for which all possible outcomes are known but the true one is not known before making the experiment Sample space A sample space is the set of all possible outcome of a random experiment.
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Examples –F–F–F–Flip or toss a coin one. The outcomes are S={ H, T} –F–F–F–Flip or toss a coin twice. The outcomes are S={ HH, HT, TH, TT} –R–R–R–Record student evaluation of a course. The outcomes are S={poor, fair, good, very good, Excellent} –F–F–F–Flip or toss a coin three times. The outcomes are S={ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT }
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The outcomes must be exhaustive ( all outcomes must be included), and they must be mutually exclusive ( no two outcomes can occur at the same time)The outcomes must be exhaustive ( all outcomes must be included), and they must be mutually exclusive ( no two outcomes can occur at the same time) Example when tossing a die one the following are exhaustive outcomes 1, 2, 3, 4, 5,Example when tossing a die one the following are exhaustive outcomes 1, 2, 3, 4, 5, 6 not
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Requirement of probabilities Given a sample spaceGiven a sample space S = { O 1, O 2, …,O k } The probabilities assigned to the outcomes must satisfy the following 1.2.
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Events An Event is a subset of the sample space SAn Event is a subset of the sample space S Example: In tossing a die one, let A be the event of having an outcome that is an even number and B be the event of having an outcome that is at least 3, and C the event of having a number greater than 5.Example: In tossing a die one, let A be the event of having an outcome that is an even number and B be the event of having an outcome that is at least 3, and C the event of having a number greater than 5. S = { 1, 2, 3, 4, 5, 6 } and A = { 2, 4, 6 } B = { 3, 4, 5, 6 } A = { 2, 4, 6 } B = { 3, 4, 5, 6 } C = { 6} C = { 6} Simple events
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Probability of an event The probability of an event is the ratio between the number of element (simple event) in the event and the total number of elements in the sample space.The probability of an event is the ratio between the number of element (simple event) in the event and the total number of elements in the sample space. Example: S = { 1, 2, 3, 4, 5, 6 } and A = { 2, 4, 6 }P(A) = 3 / 6 A = { 2, 4, 6 }P(A) = 3 / 6 B = { 3, 4, 5, 6 }P(B) = 4 / 6 B = { 3, 4, 5, 6 }P(B) = 4 / 6 C = { 6}P(C) = 1 / 6 C = { 6}P(C) = 1 / 6
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Rules of Probability Addition Rule: P(A B) = P(A) + P(B) – P(A B) P(A or B) = P(A) + P(B) if A and B are mutually exclusive Multiplication Rule: P(A B) = P(A) · P(B) if A and B are independent Complement Rule: P( A c ) = 1 – P(A)
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Example In tossing a coin 3 times, let A be the event of having exactly 2 heads, B the event of having at most 2 heads and C the event of having no heads, D no tails Then A = { HHT, HTH, THH} B = { HHT, HTH, THH, HTT, THT, TTH, TTT} C = { TTT} D ={ HHH} P(A B) = P(A) + P(B) – P(A B) = 3/8 + 7/8 - 3/8 =7/8 P(A)= 3 /8 P(B)= 7 /8P(C)=1/8 A B={ HHT, HTH, THH} P(A B)=3/8
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Rules of Probability Example A = { HHT, HTH, THH} B = { HHT, HTH, THH, HTT, THT, TTH, TTT} C = { TTT} P(A C) = P(A) + P(B) = 3/8 + 1/8 =4/8 P(C B) = P(C) + P(B) – P(C B) = 1/8 + 7/8 - 1/8 =7/8 P(A)= 3 /8 P(B)= 7 /8 P(C)=1/8 C B={ TTT} P(C B)=1/8
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Rules of Probability Example A = { HHT, HTH, THH} B = { HHT, HTH, THH, HTT, THT, TTH, TTT} C = { TTT} P(A c ) = 1 - P(A) =1 - 3/8 = 5/8 P(B c ) = 1 - P(B) =1 - 7/8 = 1/8
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Example A box containing 3 white balls, 4 black balls and 8 red balls. What is the sample space if one ball is selected at randomS = { W, B, R} What is the sample space if two ball is selected at random with replacement S = { WW, WB, WR, BB, BW, BR, RR, RW, RB} What is the sample space if two ball is selected at random with replacement S = { WB, WR, BW, BR, RW, RB}
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Rules of Probability Example A box containing 3 white balls, 4 black balls and 8 red balls. Two balls are selected at random what is the probability that the first is red and the second is white P( R and W) = P( R W) = (8/15) (3/15) P( R and W) = P( R W) = (8/15) (3/14) With Replacement Without Replacement
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Rules of Probability Example A box containing 3 white balls, 4 black balls and 8 red balls. Two balls are selected at random what is the probability that the first is white and the second is black P( W and B) = P( W B) = (3/15) (4/15) P( W and B) = P( W B) = (3/15) (4/14) With Replacement Without Replacement
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Rules of Probability Example A box containing 3 white balls, 4 black balls and 8 red balls. Two balls are selected at random what is the probability that the one is white and the one is black P(One W and One B) =P( W and B) + P( B and W) = (3/15) (4/15) + (4/15) (3/15) P(One W and One B) =P( W and B) + P( B and W) = (3/15) (4/14) + (4/15) (3/14) With Replacement Without Replacement
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Marginal and Joint probability A company ’ s employees have been classified according to age and salary, as shown in the following table: A1 B1 A2 A3 B2 B3 B P(B1(=43/100 P(B3(=26/100 P(B3(=31/100 P(A1(=35/100 P(A1 and B1(=32/100 Joint Probability Marginal Probability
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Marginal and Joint probability One employee is selected at random, what is the probability that he is between 30-45 years old A1 B1 A2 A3 B2 B3 B P(A2(= 49/100 P(A2and B2(= 18/100
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Conditional probability It is the probability of one event given that the other event has occurred. It is to know hoe two events are related We would like to know for example what is the performance of a female student will be. The probability we look for is P( performance of a student given that the student is female) Written P (B/A) where EVANT A (OCCURRED) Event B (Not Yet Occurred)
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Conditional probability Example:Example: One employee is selected at random and found to be under 30 years old, what is the probability that his salary is under $ 25,000 A1 B1 A2 A3 B2 B3 B P(A1 and B1(=32/100 P(A1 (=35/100
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Conditional probability Example:Example: One employee is selected at random and found to have a salary that between $25,000 and $45,000, what is the probability that he is over 45 yr old A1 B1 A2 A3 B2 B3 B P(A3 and B2(=10/100 P(B2 (=35/100
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BAYES ’ LAW Conditional probability is looking for the probability of a particular event when one of its causes has occurred. In many situations we may be interested in finding the probability of a specific cause for this particular event. ( A reverse thinking)Conditional probability is looking for the probability of a particular event when one of its causes has occurred. In many situations we may be interested in finding the probability of a specific cause for this particular event. ( A reverse thinking)
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BAYES ’ LAW Example: A computer chips manufactory has three production lines to produce those chips. P1 (production line 1) produces 40% of the whole production of the manufactory, P2 (production line 2) produces 35% of the whole production of the manufactory, and P3 (production line 3) produces 25% of the whole production of the manufactory. It is known that P1 produce 3% defective items, P2 produce 2% defective items, and P3 produce 1.5% defective items. One item was drawn at random and found to be defective. What is the probability that this item was produces by P2.
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P(P1) = 0.40 P(D/P1) = 0.03 P(P2) = 0.35 P(D/P2) = 0.02 P(P3) = 0.25 P(D/P3) = 0.015 The required probability is P(P2/D)
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