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Bayes Theorem Presented by: Engr. Rogelio C. Golez, Jr Mechanical Engineering Dept.

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Bayes TheoremPresented by:

Engr. Rogelio C. Golez, JrMechanical Engineering Dept.

Theorem of Total Probability

• If the events B1, B2, …, Bk constitute a partition of the sample space S such that P(Bi) ≠ 0 for I = 1, 2, …, k, then for any event A of S.

P(A) =

k

iii

k

ii BAPBPABP

11

)\()()(

EXAMPLE PROBLEMS

In certain assembly plant, three machines, B1, B2, and B3, make 30%, 45% and 25%, respectively, of the products. It is known from past experience that 2%, 3% and 2% of the products made by each machine, respectively, are defective. Now, suppose that a finished product is randomly selected. What is the probability that it is defective?

Solution:

• Consider the following:

– A: the product is defective,– B1: the product is made by machine B1,

– B2: the product is made by machine B2,

– B3: the product is made by machine B3

Solution:

P(B 1) =

0.30

P(B3 ) = 0.25

P(B2) = 0.45

B1

B2

B3

P(A\B1) = 0.02

P(A\B2) = 0.03

P(A\B3) = 0.02

Applying the rule of elimination, we can write

Let P(A) = is the probability that the product is defective.

P(A) = P(B1)P(A\B1)+P(B2)P(A\B2)+P(B3)P(A\B3)

P(A) = (0.30)(0.02) + (0.45)(0.03) + (0.25)(0.02) = 0.0245

Bayes’ Rule: derivation

)()&()/(

BPBAPBAP

• Definition:Let A and B be two events with P(B) 0. The conditional probability of A given B is:

The idea: if we are given that the event B occurred, the relevant sample space is reduced to B {P(B)=1 because we know B is true} and conditional probability becomes a probability measure on B.

Bayes’ Rule: derivation

can be re-arranged to:

)()/()&( BPBAPBAP

)()/()&( )(

)&()/( APABPBAPAPBAPABP

)()()/()/(

)()/()()/()()/()&()()/(

BPAPABPBAP

APABPBPBAPAPABPBAPBPBAP

)()&()/(

BPBAPBAP

and, since also:

Bayes’ Rule:

)()()/()/(

BPAPABPBAP

From the “Law of Total Probability”

OR

)(~)~/()()/()()/()/(

APABPAPABPAPABPBAP

Bayes TheoremLet {B1, B2, …, Bn} be a subset of the sample space S of an

experiment. If for i = 1, 2, …, n, P (Bi) > 0, then for any event A of S with P

(A) > 0,

P (Bk | A)=

P(B) is the prior probability and P(B | A) is the posterior probability

)()|(...)()|()()|()()|(

2211 nn

kk

BPBAPBPBAPBPBAPBPBAP

In-Class Exercise

• If HIV has a prevalence of 3% in San Francisco, and a particular HIV test has a false positive rate of .001 and a false negative rate of .01, what is the probability that a random person who tests positive is actually infected (also known as “positive predictive value”)?

Answer: using probability tree

______________ 1.0

P(test +)=.99

P(+)=.03

P(-)=.97

P(test - = .01)

P(test +) = .001

P (+, test +)=.0297

P(+, test -)=.003

P(-, test +)=.00097

P(-, test -) = .96903P(test -) = .999

A positive test places one on either of the two “test +” branches. But only the top branch also fulfills the event “true infection.” Therefore, the probability of being infected is the probability of being on the top branch given that you are on one of the two circled branches above.

%8.9600097.0297.

0297.)(

)&()/(

testPtruetestPtestP

Answer: using Bayes’ rule

%8.96)97(.001.)03(.99.

)03(.99.)()/()()/(

)()/()/(

truePtruetestPtruePtruetestPtruePtruetestPtesttrueP

• A box contains seven red and thirteen blue balls. Two balls are selected at random and are discarded without their colors being seen. If a third ball is drawn randomly and observed to be red, what is the probability that both of the discarded balls were blue?

• Solve P (BB | R) =

=

)()|()()|()()|()()|(

RRPRRRPBRPBRRPBBPBBRPBBPBBRP

46.0

19021*

185

19091*

186

9539*

187

9539*

187

An auto insurance company insures drivers of all ages. An actuary compiled the following statistics on the company’s insured drivers:

A randomly selected driver that the company insures has an accident.Calculate the probability that the driver was age 16-20.

Solution

An insurance company issues life insurance policies in three separate categories: standard, preferred, and ultra-preferred. Of the company’s policyholders, 50% are standard, 40% are preferred, and 10% are ultra-referred. Each standard policyholder has probability 0.010 of dying in the next year, each preferred policyholder has probability 0.005 of dying in the next year, and each ultra-preferred policyholder has probability 0.001 of dying in the next year. A policyholder dies in the next year.

What is the probability that the deceased policyholder was ultra-preferred?

Upon arrival at a hospital’s emergency room, patients are categorized according to their condition as critical, serious, or stable. In the past year:(i) 10% of the emergency room patients were critical;(ii) 30% of the emergency room patients were serious;(iii) the rest of the emergency room patients were stable;(iv) 40% of the critical patients died;(vi) 10% of the serious patients died; and(vii) 1% of the stable patients died.

Given that a patient survived, what is the probability that the patient was categorized as serious upon arrival?

A health study tracked a group of persons for five years. At the beginning of the study, 20% were classified as heavy smokers, 30% as light smokers, and 50% as nonsmokers. Results of the study showed that light smokers were twice as likely as nonsmokers to die during the five-year study, but only half as likely as heavy smokers. A randomly selected participant from the study died over the five-year period.

Calculate the probability that the participant was a heavy smoker.

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