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

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# Section 5.3: Independence and the Multiplication Rule Section 5.4: Conditional Probability and the General Multiplicat - PowerPoint PPT Presentation

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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Presentation Transcript
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!)
• 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)=

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 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

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?

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)=?

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

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