Expert: Josh - 9/5/2009

Question
Pl. solve the problem of the following:-

Q:- A bank has a test designed to establish the credit rating of a loan application. If the persons, who default (D), 90% fail the test (F). Of the persons, who will repa the bank (ND), 5 % fail the test. Furthermore, it is given that 4% of the population is not worthy of credit (i.e. defaulters). Given that someone failed the test, what is the probability that he actually will default (When given the loan)?

Answer
Manoranjan,

The discrete events are represented by the following:
F = Fail test, !F = Not fail,
D = Default,   !D = Not default

Before we analyze the problem, let us develop some notations.
P(X|Y) represents the conditional probability that X happens given the status Y.
P(X,Y)=P(Y,X) represents the joint probability that X AND Y are simultaneously true.

For example,
P(F|D) is the probability that a person fails the test given that the person will default.
P(D,F) is the probability that a person both defaults and fails the test.

Interpreting the information, we have

Probability of correct prediction of bad loans:
P(F|D)=0.9 (viz., "Fail test" knowing person "will default") ...[1]

Probability of false alarm:
P(F|!D)=0.05 (i.e., "Fail test" even though person "will not default") ...[2]

From prior knowledge, we know that 4% of population defaults,
P(D)=0.04 ...[3]

Using the complementary property, we reason
Probability of false negative (misdetection):
P(!F|D)=1-P(F|D)=0.1 (i.e., "Pass test" even though person "will default") ...[4]

Similarly, probability of correct prediction of good loans:
P(!F|!D)=1-P(F|!D)=0.95 ...[5]

OBJECTIVE:
What we need to find, is the probability that one "defaults" given a "failed test", P(D|F).
Using Bayes rule, P(D|F)=P(D,F)/P(F) .....[6]

ONE APPROACH:
Here is one technique that I propose. You should check the details and verify the answer yourself.

From a Venn diagram,

| In the figure below, two rectangles are supposed to intersect.
| Use area ABDC to denote set D, area EGIJ to denote set F.
| Area ABFEHC corresponds to P(D,!F)
| Area FGJIHD corresponds to P(!D,F)
| Intersection EFDH corresponds to P(D,F)

A--------B
|........|
|....E---F-----G
|....|...|.....|
|....|...|.....|
C----H---D.....|
.....|.........|
.....|.........|
.....|.........|
.....I---------J

P(D,!F)=P(!F,D)
      =P(!F|D)*P(D)
      =0.1*0.04 using [3]&[4]
      =0.004                  ...[7]

From the Venn diagram, visually, we can verify P(D,F)=P(D)-P(D,!F). Both P(D) and P(D,!F) are now known.
P(D,F)=0.04-0.004=0.036

Again, P(F,!D)=P(F|!D)*P(!D)
             =0.05*0.96 using [2]&[3]
             =0.048

From the Venn diagram, visually, we can verify P(F)=P(F,D)+P(F,!D).
P(F)=0.036+0.048=0.084          ...[8]

Finally, substituting [7] and [8] in [6], P(D|F)=P(D,F)/P(F)=0.036/0.084 (ans: approx. 0.4285)