Download Friday 2/21: Sections 5.3 and 5.4

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)
P(A|B)=
𝑃(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 5 heads?
A.
B.
C.
D.
0
1/25
1/10
(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 0.03.
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 | B 
P  A  B  P(test  diseased ) P(test   diseased )
0.0194



 0.2836
P( B)
P(test )
P( false)  P(true) 0.0490  0.0194
P(A and B) = P(A)∙P(B|A )
• “General Multiplication Rule”
• Algebraically equivalent to
𝑃(𝐴∩𝐵)
P(A|B)=
𝑃(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.
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