Special Type of Conditional Probability Bayes’ Theorem Special Type of Conditional Probability

Recall- Conditional Probability P(Y  T  C|S) will be used to calculate P(S|Y  T  C) P(Y  T  C|F) will be used to calculate P(F|Y  T  C) HOW????? We will learn in the next lesson? BAYES THEOREM

Definition of Partition Let the events B1, B2, , Bn be non-empty subsets of a sample space S for an experiment. The Bi’s are a partition of S if the intersection of any two of them is empty, and if their union is S. This may be stated symbolically in the following way. 1. Bi  Bj = , unless i = j. 2. B1  B2    Bn = S.

Partition Example S B1 B2 B3

Example 1 The University of Nizwa wants to hold an outdoor activity next Saturday. Past experience indicates that the probability of a successful activity is 60%, if it does not rain. This drops to 30% if it does rain on Saturday. A phone call to the weather bureau finds an estimated probability of 20% for rain. What is the probability that UN will have a successful activity?

Example 1 Events R- rains next Saturday N -does not rain next Saturday. A -activity is successful U- activity is unsuccessful. Given P(A|N) = 0.6 and P(A|R) = 0.3. P(R) = 0.2. In addition we know R and N are complementary events P(N)=1-P(R)=0.8 Our goal is to compute P(A).

Using Venn diagram –Method1 S=RN Event A is the disjoint union of event R  A & event N  A A R N P(A) = P(R  A) + P(N  A)

P(A)- Probability that you have a Successful Sale We need P(R  A) and P(N  A) Recall from conditional probability P(R  A)= P(R )* P(A|R)=0.2*0.3=0.06 Similarly P(N  A)= P(N )* P(A|N)=0.8*0.6=0.48 Using P(A) = P(R  A) + P(N  A) =0.06+0.48=0.54 Exercise: use a tree structure to reach the same result.

Let us examine P(A|R) Consider P(A|R) S=RN R N A Consider P(A|R) The conditional probability that activity is successful given that it rains Using conditional probability formula

Tree Diagram-Method 2 Probability Conditional Event Saturday R N Bayes’, Partitions Tree Diagram-Method 2 Probability Conditional Event Saturday R N A R  A 0.20.3 = 0.06 A N  A 0.80.6 = 0.48 U R  U 0.20.7 = 0.14 U N  U 0.80.4 = 0.32 0.2 0.8 0.7 0.3 0.6 0.4 P(R ) P(A|R) P(N ) P(A|N) *Each Branch of the tree represents the intersection of two events *The four branches represent Mutually Exclusive events

Method 2-Tree Diagram Using P(A) = P(R  A) + P(N  A) =0.06+0.48=0.54

Extension of Example1 Consider P(R|A) The conditional probability that it rains given that activity is successful How do we calculate? Using conditional probability formula = = 0.1111 *show slide 7

Example 2 In a recent OMANI article, it was reported that light trucks, which include AWD, pick-up trucks and minivans, accounted for 40% of all personal vehicles on the road in 2002. Assume the rest are cars. Of every 100,000 car accidents, 20 involve a fatality; of every 100,000 light truck accidents, 25 involve a fatality. If a fatal accident is chosen at random, what is the probability the accident involved a light truck?

Example 2 Events C- Cars T –Light truck F –Fatal Accident N- Not a Fatal Accident Given P(F|C) = 20/10000 and P(F|T) = 25/100000 P(T) = 0.4 In addition we know C and T are complementary events P(C)=1-P(T)=0.6 Our goal is to compute the conditional probability of a Light truck accident given that it is fatal P(T|F).

Goal P(T|F) Consider P(T|F) S=CT Consider P(T|F) Conditional probability of a Light truck accident given that it is fatal Using conditional probability formula F C T

P(T|F)-Method1 Consider P(T|F) Conditional probability of a Light truck accident given that it is fatal How do we calculate? Using conditional probability formula = = 0.4545

Tree Diagram- Method2 Probability Conditional Event Vehicle C T F C  F 0.6 0.0002 = .00012 F T  F 0.40.00025= 0.0001 N C  N 0.6 0.9998 = 0.59988 N T N 0.40.99975= .3999 0.6 0.4 0.9998 0.0002 0.00025 0.99975

Tree Diagram- Method2 = = 0.4545

Partition S B1 B2 B3 A

Law of Total Probability Let the events B1, B2, , Bn partition the finite discrete sample space S for an experiment and let A be an event defined on S.

Law of Total Probability

Bayes’ Theorem Suppose that the events B1, B2, B3, . . . , Bn partition the sample space S for some experiment and that A is an event defined on S. For any integer, k, such that we have

Focus on the Project Recall P(Y  T  C|S) will be used to calculate P(S|Y  T  C) P(Y  T  C|F) will be used to calculate P(F|Y  T  C)

How can Bayes’ Theorem help us with the decision on whether or not to attempt a loan work out? Partitions Event S Event F Given P(Y  T  C|S) P(Y  T  C|F) Need P(S|Y  T  C) P(F|Y  T  C)

Using Bayes Theorem P(S|Y  T  C)  0.477 P(F|Y  T  C)  0.523 LOAN FOCUS EXCEL-BAYES P(F|Y  T  C)  0.523

RECALL Z is the random variable giving the amount of money, in dollars, that Acadia Bank receives from a future loan work out attempt to borrowers with the same characteristics as Mr. Sanders, in normal times. E(Z)  $2,040,000.

Decision EXPECTED VALUE OF A WORKOUT=E(Z)  $2,040,000 FORECLOSURE VALUE- $2,100,000 RECALL FORECLOSURE VALUE> EXPECTED VALUE OF A WORKOUT DECISION FORECLOSURE

Further Investigation I let Y  be the event that a borrower has 6, 7, or 8 years of experience in the business. Using the range Let Z be the random variable giving the amount of money, in dollars, that Acadia Bank receives from a future loan work out attempt to borrowers with Y  and a Bachelor’s Degree, in normal times. When all of the calculations are redone, with Y  replacing Y, we find that P(Y   T  C|S)  0.073 and P(Y   T  C|F)  0.050.

Calculations P(Y   T  C|S)  0.073 P(Y   T  C|F)  0.050 P(S|Y   T  C)  0.558 P(F|Y   T  C)  0.442 The expected value of Z is E(Z )  $2,341,000. Since this is above the foreclosure value of $2,100,000, a loan work out attempt is indicated.

Further Investigation II Let Y" be the event that a borrower has 5, 6, 7, 8, or 9 years of experience in the business Let Z" be the random variable giving the amount of money, in dollars, that Acadia Bank receives from a future loan work out attempt to borrowers with 5, 6, 7, 8, or 9 years experience and a Bachelor's Degree, in normal times. Redoing our work yields the follow results.

Similarly can calculate E(Z  ) Make at a decision- Foreclose vs. Workout Data indicates Loan work out

Close call for Acadia Bank loan officers Based upon all of our calculations, we recommend that Acadia Bank enter into a work out arrangement with Mr. Sanders.