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Chapter 6 Probability Mohamed Elhusseiny [email protected] 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 Assigning Probabilities to Events • Random Experiment A random experiment is an action or process that lead to one of several possible outcomes. – 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. Assigning Probabilities to Events • Examples – Flip or toss a coin one. The outcomes are S={ H , T} – Flip or toss a coin twice. The outcomes are S={ HH, HT, TH, TT} – Record student evaluation of a course. The outcomes are S={poor, fair, good, very good, Excellent} – Flip or toss a coin three times. The outcomes are S={ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT } Assigning Probabilities to Events • 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 not exhaustive outcomes 1 , 2 , 3 , 4, 5, 6 Requirement of probabilities • Given a sample space S = { O1, O2,…,Ok} The probabilities assigned to the outcomes must satisfy the following 1. 0 P(Oi ) 1 2. P(Oi ) 1 Events • An 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. S={1,2,3,4,5,6} and A={2,4, 6} B={3,4,5, 6} C = { 6} Simple events 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. Example: S = { 1 , 2 , 3 , 4 , 5 , 6 } and A={2,4, 6} P(A) = 3 / 6 B={3,4,5, 6} P(B) = 4 / 6 C = { 6} P(C) = 1 / 6 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) • Complement Rule: P( Ac ) = 1 – P(A) if A and B are independent Rules of Probability • 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 A B={HHT, HTH, THH} P(A)= 3 /8 Then P(A B)=3/8 A = { HHT, HTH, THH} P(C)=1/8 B = { HHT, HTH, THH, HTT, THT, TTH, TTT} P(B)= 7 /8 C = { TTT} D ={ HHH} P(A B) = P(A) + P(B) – P(A B) Rules of Probability C B={TTT} P(A)= 3 /8 A = { HHT, HTH, THH} P(C B)=1/8 B = { HHT, HTH, THH, HTT, THT, TTH, TTT} P(B)= 7 /8 P(C)=1/8 C = { TTT} • Example 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 Rules of Probability • Example A = { HHT, HTH, THH} B = { HHT, HTH, THH, HTT, THT, TTH, TTT} C = { TTT} P(Ac) = 1 - P(A) =1 - 3/8 = 5/8 P(Bc) = 1 - P(B) =1 - 7/8 = 1/8 Rules of Probability • • • • 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 random S = { 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} 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 With Replacement P( R and W) = P( R W) = (8/15) (3/15) P( R and W) = P( R W) = (8/15) (3/14) Without Replacement 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 With Replacement P( W and B) = P( W B) = (3/15) (4/15) P( W and B) = P( W B) = (3/15) (4/14) Without Replacement 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 With Replacement 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) Without Replacement Marginal and Joint probability • A company’s employees have been classified according to age and salary, as shown in the following table: B1 P(A1(=35/100 B3 B2 A1 A3 B P(A1 and B1(=32/100 A2 P(B1(=43/100 Marginal Probability Joint Probability P(B3(=26/100 Marginal and Joint probability • One employee is selected at random, what is the probability that he is between 30-45 years old P(A2and B2(= 18/100 P(A2(= 49/100 B1 B B3 B2 A1 A3 A2 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. Event B (Not Yet Occurred) EVANT A (OCCURRED) • The probability we look for is P( performance of a student given that the student is female) P( B and A) Written P (B/A) where P( B and A) P( A) P( B / A) P( B / A) P( A) Conditional probability • 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 P( B3 and A1 ) 32 / 100 32 P( B3 / A1 ) P( A1 ) 35 / 100 35 B P ( B / A) P(A1 and B1(=32/100 B1 B3 B2 A1 A3 A2 P(A1 (=35/100 Conditional probability • 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 P( A3 and B2 ) 10 / 100 10 P( A3 / B2 ) P( B2 ) 31 / 100 31 B P ( B / A) P(A3 and B2(=10/100 B1 B3 B2 A1 A2 A3 P(B2 (=35/100 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) 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. BAYES’ LAW • P(P1) = 0.40 • P(P2) = 0.35 • P(P3) = 0.25 P(D/P1) = 0.03 P(D/P2) = 0.02 P(D/P3) = 0.015 The required probability is P(P2/D) P( D) P( D / P1) P( A1) P( D / P2) P( A2) P( D / P3) P( A3) P( D) 0.03 0.40 0.02 0.35 0.015 0.25 0.02275 P( P 2 and D) 0.02 0.35 P( P 2 / D) 0.3078 P( D) 0.02275