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Conditional Probability Brian Carrico Nov 5, 2009 What is Probability? Predicting a random event – A random event is one in which individual outcomes are uncertain but the long-term pattern of many individual outcomes is predictable and every possible outcome can be described prior to its performance We can use the long-term patterns to predict individual outcomes What is Conditional Probability? In some situations current or previous conditions have an impact on the probability Examples: – – – – Weather Stock Market Genetics Card games What sort of factors can impact probability? What is the probability of rolling an even number on a fair six-sided die? – What if you’re told the roll was less than 4? – 1/2 1/3 How did you come up with that? Basic Formula for Conditional Probability P(A|B) = P(A∩B) P(B) Some Practice Female Male Dem 21 18 Rep 18 24 Ind 8 11 Total 47 53 Total 39 42 19 100 P(Female) P(Female|Democrat) P(Republican) P(Republican|Male) 47/100 = 0.47 21/39 = 0.538 42/100 = 0.42 24/53 = 0.453 Law of Total Probability If A is some event and {B1, B2, … Bn} forms a partition of the sample space then: P(A)=ΣP(A|Bi)*P(Bi) Proving P(A)=ΣP(A|Bi)*P(Bi) U{B1, B2,…, Bn} = S P(A) = P(A∩S) P(A) = P(A∩(U{B1, B2,…, Bn} )) P(A) = P(U{A∩B1, A∩B2,…, A∩Bn}) P(A) = ΣP(A∩Bi) P(A) = ΣP(A|Bi)*P(Bi) Using the Law of Total Probability Suppose you have two urns containing balls colored green and red. Urn I contains 4 green balls and 6 red balls, Urn II contains 6 green balls and 3 red balls. A ball is moved from Urn I to Urn II at random then a ball is drawn from Urn II. Find the probability that the ball drawn from Urn II is green. Urn Problem Continued Events: – – – G1=Ball transferred from Urn I to Urn II is Green R1=Ball transferred from Urn I to Urn II is Red G2=Ball drawn from Urn II is Green We want P(G2) We have – – – P(G2)= P(G2|G1)*P(G1) + P(G2|R1)*P(R1) P(G2)=(7/10)*(4/10) + (6/10)*(6/10) P(G2)=28/100+36/100=64/100 Bayes’ Rule If A is some event and {B1, B2, … Bn} forms a partition of the sample space then: P(Bj|A)= _P(A|Bj)*P(Bj) ΣP(A|Bi)*P(Bi) Using Bayes’ Rule You are tested for a disease that occurs in 0.1% of the population. Your physician tells you that the test is 99% accurate. If the test comes back positive, what is the probability that you have the disease? Events: – T=positive test D=you have the disease Test result continued Given Probabilities: – P(D)=0.001 P(T|D)=0.99 P(T|Dc)=0.01 We want P(D|T) From Bayes’ Rule we know – – – P(D|T)= P(T|D)*P(D) ___ P(T|D)*P(D)+ P(T|Dc)*P(Dc) P(D|T)= (0.99*0.001) _ (0.99*0.001)+(0.01*0.999) P(D|T)=0.09 Testing Independence If A and B are two independent events then P(A|B)=P(A) Using formulas from earlier we can see that P(A|B)=P(A∩B)=P(A) P(B) So, P(A∩B)=P(A)*P(B) A test of Independence A fair coin is tossed twice. Are the following events independent? – A= 1st toss lands heads B= 2nd toss lands heads S={HH,HT,TH,TT} P(A)=1/2 P(B)=1/2 P(A∩B)=1/4 P(A)*P(B)=1/2*1/2=1/4=P(A∩B) Homework Sources Probability Models by John Haigh 2002 Probability by Larry Leemis 2009