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45-733 PROBABILITY AND STATISTICS I Topic #1C

25 January 2000 and 27 January 2000

- Suppose we have a K-Fold partition (as in (15) of notes #1) of
our sample space,
**S**. Over the top of this partition we drop a set**B**so that**B**intersects the K partitioning sets (**A**). That is:_{1}, A_{2}, ... , A_{k}

Clearly**B**can be written as:

**B = (A**_{1}Ç B) È (A_{2}Ç B) È (A_{3}Ç B) È etc.

Now, using the identity:**P(B|A**, and rearranging:_{i}) = P(B Ç A_{i})/P(A_{i})

**P(A**. So that:_{i}Ç B) = P(A_{i})P(B|A_{i})

**P(B) = P(A**etc._{1})P(B|A_{1}) + P(A_{2})P(B|A_{2}) + P(A_{3})P(B|A_{3}) +

This allows us to state the famous**Bayes Theorem.**

Let**A**be a partition of_{1}, A_{2}, ... , A_{k}**S**such that**P(A**for all i,_{i}) > 0

and let**B**be an event such that**P(B) > 0**. Then:

**P(A**_{i}|B) = [P(A_{i})P(B|A_{i})]/ [å_{j=1,k}P(A_{j})P(B|A_{j})] =

P(A_{i}Ç B)/ [å_{j=1,k}P(A_{j}Ç B)] =

P(A_{i}Ç B)/P(B) - Example: We have three chests, each with two drawers. (The chests
are identical in every respect.) In one chest there is a silver coin in one
drawer and a silver coin in the second drawer. In a second chest there is a
gold coin in one drawer and a silver coin in the second drawer. And in the third
chest there is a gold coin in each drawer.

We randomly select a chest and open a drawer. What is the probability that the un-opened drawer contains a gold coin?

Let**A**,_{1}**A**,_{2}**A**, represent the three chests, and let_{3}**B**be the event "gold coin in the drawer". Clearly:

**P(A**=_{1})**P(A**=_{2})**P(A**= 1/3; and:_{3})

**P(B|A**= 0,_{1})**P(B|A**= 1/2,_{2})**P(B|A**= 1._{3})

Only if we chose the 3rd chest could there be a second gold coin. Hence, we need to compute:

**P(A**_{3}|B) = [P(A_{3})P(B|A_{3})]/ [å_{j=1,3}P(A_{j})P(B|A_{j})] =

= (1/3 x 1)/(1/3 x [0 + 1/2 + 1]) = 2/3. - Voting Example:
In a particular city we have three kinds of voters: Republicans,
Democrats, and Reformers,
which we represent as the events
**R**,**D**, and**J**, respectively. Let**V**be the event "voted in the last election". We are given that:

**P(R)**= .4,**P(D)**= .35, and**P(J)**= .25; and:

**P(V|R)**= .7,**P(V|D)**= .65,**P(V|J)**= .4.

We randomly select a person and learn that she*did not*vote. What is the probability that she is a Democrat?

By the rule of the complement:

**P(V**= .3,^{c}|R)**P(V**= .35,^{c}|D)**P(V**= .6.^{c}|J)

The probability we need to compute is:

**P(D|V**^{c}) = [P(D)P(V^{c}|D)]/

[P(D)P(V^{c}|D) + P(R)P(V^{c}|R)] + P(J)P(V^{c}|J)]

= (.35 x .35)/[(.35 x .35) + (.4 x .3) + (.25 x .6)] = .312 - Problem 2.83 p.63

Let**D**be the event "Disease" and**B**be the event "Positive Test". We are given in the problem statement that:

**P(D)**= .01 so that**P(D**= .99;^{c})

**P(B|D)**= .9 and**P(B**= .9.^{c}|D^{c})

We are asked to compute:

**P(D|B) = [P(D)P(B|D)]/[P(D)P(B|D) + P(D**=^{c})P(B|D^{c})]

= (.01 * .9)/[(.01 * .9) + (.99 * .1)] = .0833.

Note that**P(D|B) > P(D)**. That is, we start with aprobability of .01 that any randomly chosen person has the disease. Then we are given*prior*("postive test"). This must raise the probability that the person has the disease. This probability is our*information*probability. In the jargon of Bayesian statistics, we have*posterior*in light of the information.*revised our priors*

Also note that the probability**1 - P(D|B)**is the probability of a. This is not a costly error in this context. Far more costly is a*false positive*which in the context of disease testing means that you tell a sick person that they are not sick and send them back into the population to spread the disease. This probability is:*false negative*

**P(D|B**=^{c}) = [P(D)P(B^{c}|D)]/ [P(D)P(B^{c}|D) + P(D^{c})P(B^{c}|D^{c})]

= (.01 * .1)/[(.01 * .1) + (.99 * .9)] = .00112.

Given how costly this error is the goal of disease testing is to drive this probability to zero while keeping**P(D|B)**as high as possible. - Problem 2.93 p.64

We have five bowls, i=1, 2, 3, 4, 5, with i white and 5-i black balls in the ith bowl, respectively. The experiment consists of randomly choosing a bowl and randomly drawing two balls from the bowl. That is, we simply take two balls out of the bowl (in contrast, drawing*without replacement*means that we would take one ball out, note its color, drop it back into the bowl, and then take another ball out and note its color).*with replacement*

Part (a): What is the probability that both balls are White?

Let**A**,_{1}**A**, ...,_{2}**A**, represent the five bowls, and let_{5}**B**be the event "two White balls are drawn". Namely:

Clearly:

**P(A**=_{1})**P(A**= ... =_{2})**P(A**= 1/5; and:_{5})

**P(B|A**= 0_{1})

**P(B|A**= {[2 choose 2][3 choose 0]}/[5 choose 2] = 1/10_{2})

**P(B|A**= {[3 choose 2][2 choose 0]}/10 = 3/10_{3})

**P(B|A**= {[4 choose 2][1 choose 0]}/10 = 6/10_{4})

**P(B|A**= [5 choose 2]/10 = 1_{5})

Hence:

**P(B) =[å**= (1/5)( 0 + 1/10 + 3/10 + 6/10 +1) = 2/5_{j=1,5}P(A_{j})P(B|A_{j})]

Part (b): Given that both balls are White, what is the probability that they were drawn from bowl 3?

This is the probability:

**P(A**_{3}|B) = P(A_{3}Ç B)/P(B) = [P(A_{3})P(B|A_{3})]/P(B)] =

(1/5 x 3/10)/(2/5) = 3/20