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Section 5.3: Independence and the Multiplication Rule Section 5.4: Conditional Probability and the General Multiplicat

Section 5.3: Independence and the Multiplication Rule Section 5.4: Conditional Probability and the General Multiplication Rule. Conditional Probability “Probability of event A given event B” Event A happens given that event B will or has occurred. P(A|B) (Not a division sign!)

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Section 5.3: Independence and the Multiplication Rule Section 5.4: Conditional Probability and the General Multiplicat

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  1. Section 5.3: Independence and the Multiplication Rule Section 5.4: Conditional Probability and the General Multiplication Rule

  2. Conditional Probability • “Probability of event A given event B” • Event A happens given that event B will or has occurred. • P(A|B) (Not a division sign!) • P(snow|December) is very different from P(snow|July) and P(snow) • Roll two dice: • P(12)=1/36 • P(12|first roll 6) = 1/6 • P(12| first roll 5) = 0 • P(A|B)=

  3. Independence • Event A and B are independent if the outcome of one event does NOT affect the probability of the other event • If A and B are independent then: • P(A and B) = P(A) P(B) “multiplication rule for independent events” • P( A |B ) = P( A ) • P( B | A ) = P( B ) • If one of the above is true, so are the other two. • If two events are not independent, they are dependent.

  4. The multiplication rule generalizes: If events A, B, C, etc. are independent then P( A and B and C … ) = P(A) P(B) P(C)…..

  5. If 5 fair coins are tossed, what is the probability of getting 5 heads? • 0 • 1/25 • 1/10 • (1/2)5

  6. If 5 fair coins are tossed, what is the probability of getting at least one tail? • 0 • (1/2)1+(1/2)2+(1/2)3+(1/2)4+ (1/2)5 • 1-(1/2)5 D. (1/2)5

  7. Suppose the probability of a random bank customer defaulting on a loan is 0.2. Also suppose the probability of a random bank customer who has a FICO score of 700 or higher is 0.03. Question: Are FICO scores and likelihood of defaulting on a loan dependent or independent? Why or why not?

  8. More advanced tree diagram: Prevalence of disease: P(disease)=0.02 Sensitivity of test: P(test positive given person diseased)=0.97 Specificity of test: P(test negative given person not diseased)=0.95

  9. So given: P(disease)=0.02 P(test positive given person diseased)=0.97 P(test negative given person not diseased)=0.95 Determine probability of a person who tested positive actually not having the disease. That is, P(not diseased | person tested positive)=?

  10. Probability a random person ends up in this column (sum=1)

  11. P(A and B) = P(A)∙P(B|A ) • “General Multiplication Rule” • Algebraically equivalent to P(A|B)= • Allows us to multiply down branches in tree diagrams

  12. Suppose Team A and B will play each other in a “Best of 3” series. (First team to win two games wins the series.) Also suppose the probability of the superior Team A winning each game is 0.6. Use a tree diagram to determine the probability of Team B winning the series. • P(Team B wins series)=0 • P(Team B wins series)=0.16 • P(Team B wins series)=0.352 • P(Team B wins series)=0.4

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