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Conditional Probability
Brian Carrico
Nov 5, 2009
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
What is Conditional Probability?
In some situations current or previous conditions have an impact on the probability
Examples:– Weather– Stock Market– Genetics– Card games
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?
Basic Formula for Conditional Probability
P(A|B) = P(A∩B) P(B)
Some Practice
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
Dem Rep Ind Total
Female 21 18 8 47
Male 18 24 11 53
Total 39 42 19 100
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)
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)
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.
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
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)
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
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
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)
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)
Homework
Sources
Probability Models by John Haigh 2002
Probability by Larry Leemis 2009