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Conditional Probability

Conditional Probability. Brian Carrico Nov 5, 2009. What is Probability?. Predicting a random event

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Conditional Probability

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  1. Conditional Probability Brian Carrico Nov 5, 2009

  2. 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

  3. What is Conditional Probability? • In some situations current or previous conditions have an impact on the probability • Examples: • Weather • Stock Market • Genetics • Card games

  4. What sort of factors can impact probability? • What is the probability of rolling an even number on a fair six-sided die? • 1/2 • What if you’re told the roll was less than 4? • 1/3 • How did you come up with that?

  5. Basic Formula for Conditional Probability • P(A|B) = P(A∩B) • P(B)

  6. 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 Some Practice

  7. 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)

  8. 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)

  9. 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.

  10. 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

  11. 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)

  12. 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

  13. 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

  14. 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)

  15. 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)

  16. Homework

  17. Sources • Probability Models by John Haigh 2002 • Probability by Larry Leemis 2009

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