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Topic 2: Intro to probability CEE 11 Spring 2002 Dr. Amelia Regan These notes draw liberally from the class text, Probability and Statistics for Engineering and the Sciences by Jay L. Devore, Duxbury 1995 (4th edition) This topic also draws from Probability Concepts in Engineering Planning and Design: Volume I - Basic Principles by A. H-S. Ang and W. H.Tang, John Wiley & Sons, 1975 definitions The set of all outcomes in a probabilistic problem is collectively a sample space. Each of the individual possibilities is a sample point. An event is defined as a subset of the sample space. Sample space may be discrete or continuous. A discrete sample space may be finite or countably infinite. Examples: {H,T}, {1,2,3,4,5,6}, {1/2,1/4,1/8,…1/n} Class Exercise Problem 3.6, slightly modified. Starting at a fixed time, each car entering an intersection is observed to see whether it turns left (L), right (R) or goes straight ahead (S). The experiment terminates as soon as a car is observed to turn left. Let the random variable X be equal to the number of cars entering the intersection before a car turns left. What are the possible values for X? List 5 outcomes and their associated X values. definitions If the occurrence of one event precludes the occurrence of another event the two are said to be mutually exclusive. If two events A and B are mutually exclusive then: P(A B)=0 Two or more events are said to be collectively exhaustive if the union of all of these events constitute the underlying sample space. some basic properties of probability For any event A, P(A) 0. (no negative probabilities) Let S represent the sample space, P(S) = 1 (probabilities sum to one) if A1, A2, ...,An is a finite collection of mutually exclusive events, then n P( A1 A2 ... An ) P( Ai ) i 1 (the probability that at least one will occur is the sum of the individual probabilities of each occurring) some basic properties of probability For any event A P(A) = 1-P(A) Illustrate this with a Venn Diagram here some basic properties of probability For any two events A and B P(A B) = P(A)+P(B)-P(A B) Illustrate this with a Venn Diagram here some basic properties of probability Similarly P(A B C) = P(A)+P(B)+P(C) -P(A B)-P(A C)-P(B C) +P(A B C) Illustrate this with a Venn Diagram here some basic properties of probability For any events A and B ________ A B = A B de Morgan's Rule ________ P(A B) = P(A B) Illustrate this with a Venn Diagram here conditional probability For any two events A and B with P(B) > 0, the conditional probability of A given that B has occurred is defined by P( A B) P( A | B) P( B) Note that the multiplication rule for P(A intersection B) follows directly P( A B) P( A | B) P( B) Example In a game show a contestant is asked to pick from one of three doors. Behind one of the doors is a new car which he gets to keep if he selects that door. 1 P ( D1 ) P ( D2 ) P ( D3 ) 3 The contestant picks door number 1. One of the other two doors is selected at random and is opened (say door 2). The car is not behind door 2. Now what are the chances that the car is behind door number 1? 1 P( D1 D2 ) 3 1 P( D1 | D2 ) P( D2 ) 2 2 3 The “Monty Hall” Problem This is actually a well known problem. In a game show a contestant is asked to pick from one of three doors. Behind one of the doors is a new car which he gets to keep if he selects that door. 1 P ( D1 ) P ( D2 ) P ( D3 ) 3 The contestant picks door number 1. The host (Monty Hall), who knows where the prize is, opens a door behind which there is no prize. The contestant is offered a chance to switch his choice. Should he switch? Bayes’ Theorem The multiplication rule P( A B) P( A | B) P( B) leads directly to Bayes’ Theorem which is used to compute posterior probabilities from given prior probabilities. First we need the law of total probability... Bayes’ Theorem Let A1, A2, ..., An be a collection of n mutually exclusive and collectively exhaustive events. Then for any other event B for which P(B) > 0, The law of total probability states that P( B) P( B | A1 ) P( A1 ) P( B | A2 ) P( A2 ) ... P( B | An ) P( An ) n P( B | Ai ) P( Ai ) i 1 Bayes’ Theorem B A1 A2 … An Remember the diagram associated with the law of total probability The rectangular region represents the universe which is divided into non-overlapping areas which together cover the whole region. The probability that B occurs the sum of the probabilities that B occurs with any of the mutually exclusive and collectively exhaustive events. Bayes’ Theorem Let A1, A2, ..., An be a collection of n mutually exclusive and collectively exhaustive events with P(Ai) > 0 for i=1,...,n. Then for any other event B for which P(B) > 0 P( Ak B) P( B | Ak ) P( Ak ) P( Ak | B) n k 1, 2,...n P( B) P( B | Ai ) P( Ai ) i 1 The second form follows directly from the first by using the multiplication rule in the numerator and the law of total probability in the denominator Example (law of total probability) Assume that 1 in 10,000 adults is inflicted with a rare disease. A diagnostic test is developed – the test is 99% accurate (1% false negatives and 1% false positives) If a person tests positive – what is the probability she has the disease? – Less than 10%? P( D P) P( D P) P( P) P( P | D) P( D) P( P | D ) P( D ) (0.001)(0.99) 0.00099 (0.99)(0.001) (0.01)(0.999) 0.00999 0.00999 0.00099 0.09016 0.01098 P( D | P) independence Two events A and B are independent if P( A | B) P( A) They are dependent otherwise independence Remember that in general P( A B) P( A | B) P( B) However, if A and B are independent then P( A B) P( A) P( B) In fact, A and B are independent if and only if the above is true independence Events A1, A2, ...., An are mutually independent if for every k (k = 2,3,...,n) and every subset of indices i1, i2, i3,...,in, P( Aii Ai2 ... Ain ) P( Aii )P( Ai2 )P( Ain ) independence Question: If two events are mutually exclusive can they also be independent? Class Exercise A construction company has the opportunity to bid on three jobs (A,B and C). From historical data the company believes its changes of winning each job is 0.20, 0.50 and 0.10. Its chances of winning all three jobs is 0.05. The probability that it will win both A and B is 0.10. For A and C and B and C the probabilities are 0.07 and 0.06. Are the events that the construction company is awarded jobs A, B and C independent? Why or why not? Class Exercise A construction company has the opportunity to bid on three jobs (A,B and C). From historical data the company believes its changes of winning each job is 0.20, 0.50 and 0.10. Its chances of winning all three jobs is 0.05. The probability that it will win both A and B is 0.10. For A and C and B and C the probabilities are 0.07 and 0.06. Calculate the following probabilities: The company wins none of the jobs The company wins only job A The company wins only one of the jobs Hint -- draw and label a Venn Diagram Class Exercise A construction company has the opportunity to bid on three jobs (A,B and C). From historical data the company believes its changes of winning each job is 0.20, 0.50 and 0.10. Its chances of winning all three jobs is 0.05. The probability that it will win both A and B is 0.10. For A and C and B and C the probabilities are 0.07 and 0.06. Calculate the following conditional probabilities: A given B C given A C given A and B