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

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Presentation transcript:

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

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.

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)…..

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

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

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 Question: Are FICO scores and likelihood of defaulting on a loan dependent or independent? Why or why not?

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

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)=?

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

P(A and B) = P(A)∙P(B|A )

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. A.P(Team B wins series)=0 B.P(Team B wins series)=0.16 C.P(Team B wins series)=0.352 D.P(Team B wins series)=0.4