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