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BUSINESS STATISTICS (PART- 19)
Probability-5
[Conditional probability, Total probability Law,
Bayes’ Rule]
1. INTRODUCTION
Hello viewers, today we shall continue our discussion on probability
theory. In last two lectures we have discussed the definitions of
probability and learned how to calculate the probability of some
simple events using the counting methods namely:
 fundamental principal of counting, and
 Permutations and combinations.
For example, what is the probability that a given card which is
selected randomly from a deck of 52 cards is a face card? And we know
that the probability of this event is 12/52. But suppose some additional
condition is given about this event. For example, that the selected
card is of red colour then the probability of this event will be different
and because we have impose some condition or we provided some
additional information on this event and this gives rise to the
conditional probability. So far, we have discussed two types of
probabilities: simple and joint i.e. P(A) and P(A∩B) or P(A or B). In this
lecture we are going to discuss conditional probability and the Bayes’
Theorem/ Bayes’ Rule, which involves conditional probabilities. We
shall then discuss its applications.
2. WHY CONDITIONAL PROBABILITY?
Now, the question arises why conditional probabilities. Sometimes the
simple probability changes in the light of some related information or
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some condition is imposed on the event under consideration .That
gives rise to concept of conditional probability and its laws.
Conditional and joint probabilities under statistical dependence are
more involved than simple probabilities. For example, during a study
of auto accidents, the highway Safety council found that 60% of all
accidents occur at night, 52% are alcohol-related, and 37% occur at
night and are alcohol related. Then the probability that an accident
occurred at night is a simple probability but this is not enough. More
appropriate is to know the “probability that an accident occurred at
night given that it was alcohol related”.
This is not simple probability. It is a conditional probability or
Probability of an event under certain condition.
3. CONDITIONAL PROBABILITY
Now, in order to understand how to compute conditional probabilities,
let us look at this example:
 A Box contains 4 Red & 3 Green identical marbles except for the
color.
 Two marbles are selected at random one by one, without
replacement and taking order into account.
Then we define event:

A= the first selected marble is Red.

B= the second selected marble is Green.
Now, in this experiment we know that there are 42 possible outcomes
of the experiment i.e. the sample space consists 42 elementary events.
This we can obtain by using the permutations i.e. 7P2. Now here the
event A which tells that the first selected marble is red. If, for
example, the red marbles are denoted by say R1, R2, R3 and R4, and the
green marbles are denoted by G1, G2, G3 then this subset A will consist
of the elementary events like: R1R2, R1R3, R1R4 . . . R1G3.
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This way there will be 24 elementary events in subset A. Therefore,
P(A)=24/42=4/7.
Now, the subset B will consist of the elementary events with the
second selected marble is of green color i.e. it will consist of the
elementary events like: R1G1, R1G2 or G1G2 or G1G3 something like
that. So the (A∩B), when we find out it will consist only 12 elementary
events and the P(A∩B)= 12/42, using our m/n definition of probability.
So,
P(A)= 4/7
and
P(A∩B)= 2/7.
Then P(BIA) is the conditional probability. “B given A” we denote by
the notation B a vertical bar (I) and then event A i.e. B|A. So this P( B
| A) we call Probability of B given A and it is a conditional probability.
P (A  B)
This is equal
to
and this probability is 2/4. So, what is the
P(A)
main idea here in the conditional probability? That is, we put a
condition on an event, here we put a condition on event B and then we
find out the probability of B in the presence of event A .Then we are
restricting the sample space for this event. So for the event B given A,
the sample space is the subset A and not the original sample space S.
So, now we can give a formal definition of conditional probability that:
P(BIA)= P (A  B) ,
or, P(AIB)=
P(B)
P (A  B) ,
P(A)
provided P(A)≠0
provided P(B)≠0
Say, this restriction is necessary to define the conditional probability.
That is P(A) should not be equal to zero in case of P(BIA). And in case
of P(AIB) there is a condition that P(B) should not be equal to zero. If
this condition is not satisfied what will happen? Suppose P(A) is zero
then P(BIA) will be infinity. And we know that probability of an event
cannot be greater than one. So, it is necessary to write this condition
when we are defining the conditional probabilities or this can also be
written as:
P(A∩B)= P(A) P(BIA)= P(B) P(AIB)
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Now, we should be very careful about the notation for the conditional
probability. In conditional probability we use notation like P(BIA) or
P(AIB). Here in BIA or AIB a vertical bar is used and, it should not be
written as B/A (B slanted bar A) or B , as these have no meaning when
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A and B are events.
Now, if the two events A and B are independent. Then
P(BIA)=P(B) , P(A)≠0 and
P(AIB)=P(A) , P(B)≠0
As we know that when two events A and B are independent then
P(A∩B)= P(A) P(B)
When we put this condition in the definition of the conditional
probability then we simply get these two results, i.e.,
P(BIA)=P(B) , P(A)≠0 and
P(AIB)=P(A) , P(B)≠0.
4. TOTAL PROBABILITY
Now, we shall define the total probability. In order to understand it we
consider an Example: Dr. Jay –a medical practitioner
Dr. Jay is not a very good medical practitioner. If a patient goes to him
 The chance that he will diagnose the patient’s symptoms
properly is 30 %.
 Even if Dr. Jay’s diagnose is correct his treatment is such that
the chance of the patient dying is 60%.
 and if diagnose is wrong the chance of patient dying during
treatment is 95%.
Here we denote:
A1 : be the event of correct diagnosis. And
A2 : be the event of wrong diagnosis. And
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B : the patient of Dr. Jay dying.
What else is given in this example is the conditional probabilities i.e.
P(BIA1). That is, the probability of dying the patient when he was
correctly diagnosed. And it is already given as 0.6. And the P(BIA2) i.e.,
probability of dying the patient when he was wrongly diagnosed by Dr.
Jay is 0.95. So,
P(A1) and P(A2 ) these are the simple probabilities.
And
P(BIA1)and P(BIA2) these are our conditional probabilities.
Now, when these probabilities are given how we obtained the P(B) i.e.
probability that a patient dies during the treatment. And this
probability can be obtained as:
P(B∩A1)+ P(B∩A2), using the additive law of probability. And,
P(B∩A1)=P(BIA1).P(A1) and P(B∩A2)=P(BIA2).P(A2).
And this, on simplification gives the P(B)= 0.845. So, here this P(B)
given in equation (1) or in equation (2) are known as total probability
law. So, total probability law is expressed in terms of the conditional
probabilities.
5. BAYES’ RULE
Now, we are going to discuss a very important result on conditional
probability i.e. known as the Bayes’ Rule or the Bayes’ theorem. Now,
in the example of Dr. Jay, if he want to find out what is the P(A2IB) or
what is the P(A1IB)? That is, what is the probability that the patient
was wrongly diagnose in the presence that the patient had died, i.e.
P(A2IB). Now you remember here we also know the P(BIA2).
We are already given the P(BIA2). But here we want to find out P(A2IB).
So there is a difference. P(BIA2) is different than P(A2IB), and the
P(A2IB) we can obtained by this formula. That is,
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P( A2 B) 
P(B A 2 ) P(A 2 )
P(B A1 )P(A 1 )  P(B A 2 )P(A 2 )
   (1)
When we substitute the values of all these conditional and simple
probabilities on the R.H.S. of this formula (1), we get the P(A2IB) as
0.787. Now here I just told that there are two types of conditional
probabilities: P(BIA1) or P(BIA2) and also we obtained P(A2IB). So, the
P(A2IB) is obtained with the help of the conditional probabilities of
(BIA1) and (BIA2). Therefore, the P(BIA1) and P(BIA2) are known as a
prior probabilities, because unless these probabilities are given we
cannot calculate P(A2IB). And the P(A2IB) is called the posterior
probability which we obtained latter with the help of prior conditional
probabilities. So, formula given in Equation (1) connects prior and
posterior probabilities and it is known as Bayes’ Rule.
So, in general, let a sample space S partitioned into mutually exclusive
exhaustive events A1, A2, . . . ,Ak and let B be any other event in the
same sample space. i.e.
S
A1
A2
Ak
B
Here, the sample space S is partitioned into K different subsets
A1, A2, . . . ,Ak. And these subsets are mutually exclusive i.e. there is
nothing common between A1 & A2 or A1 & A4. And let B be any arbitrary
event.
So in our Dr. Jay’s example our subsets A1 & A2 were mutually
exclusive, because either a patient will be diagnose correctly or it can
be diagnose wrongly.
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Their (A1∩A2) was a null set i.e. they were mutually exclusive. Their
the sample space was partitioned only in two subsets, but when we
generalised it we say that this sample space is partitioned into ksubsets A1, A2, . . . ,Ak and B is any arbitrary event. So we see that
A1∩B or A2∩B are not the null sets. So the total probability law:
P(B)=P(BIA1)P(A1)+P(BIA2)P(A2)+……+P(BIAk)P(Ak)
And, the Bayes’’ Rule is:
P( Ai B) 
P( B | Ai ) P( Ai )
P( B | A1 ) P( A1 )      P( B | Ak ) P( Ak )
That is, in the Bayes’ Rule this P(BIAi)P(Ai) is divided by the total
probability i.e. P(B) and of course with the condition that:
P(Ai)≠0, for all i=1,2…….k.
That is, none of the probabilities of the events Ai will be zero. If it is
any one of them is zero, then this rule will not work.
Here, P(BIAi) are called Prior probabilities as we have just discussed &
P(AiIB) are Posterior probabilities.
6. APPLICATION OF BAYES’ RULE
Let us see how the Bayes’ Rule can be used to solve some complex
problem on probabilities. Here let us looks an example:
Martin Coleman, he is credit Manager for a nationalized Bank. He
knows that the company uses three methods to encourage collection of
delinquent accounts. From past collection records he learns that 70%
of the accounts are called on personally, 20% are phoned and 10% are
sent a letter.
The probabilities of collecting an overdue amount from an account
with the three methods are respectively 0.75, 0.60 and 0.65. Mr.
Coleman has just received payment from a past- due account.
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Now the question is:
What is the probability that if Mr. Coleman has just received
payment from a past due account then, it was due to personal call?
This probability perhaps he wants to know that which method of
collecting the money from the overdue account is effective. So, in
order to calculate this probability let us define events A1, A2 and A3.
Here
Event A1 : delinquent accounts are called on personally
Event A2 : delinquent accounts are called on phone, and
Event A3 : delinquent account holders are sent letters.
And B : Collection of overdue amount.
Now, here we observe that there is nothing common between events A1
and A2; and A1, A3; and A2, A3 i.e. these events are mutually exclusive.
That is:
(A1∩A2) is a null set,
(A1∩A3) is also a null set, and
(A2∩A3) is also a null set.
Also A1, A2 and A3, they exhaust the sample space completely. So, the
primary condition of the Bayes’ theorem is satisfied. And, now we are
given that:
P(A1)=0.7, P(A2 )=0.2, P(A3 )=0.1
We are also given the conditional probabilities:
P(BIA1)=0.75, and
P(BIA2)=0.60, and
P(BIA3)=0.65.
So, by using the total probability law, we get
P(B)=0.75x0.7+0.6x0.2+0.65x0.1=0.710
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And, using the formula of the Bayes Rule, we get
P(A1IB)=0.75x0.7/0.710=0.739.
Similarly, the Bayes’ Rule provide
P(A2IB)=0.169, and P(A3IB)=0.092
That is, the probability that the past-due a/c called on phone, and
through letter respectively.
So, here these probabilities: P(A1IB), P(A2IB) and P(A3IB) are the
posterior probabilities. And these probabilities are obtained using the
Bayes’ Rule which uses the conditional probabilities like: P(BIA1),
P(BIA2) and P(BIA3) i.e. the prior probabilities.
7. SUMMARY
So, in this part of the lecture today we have discussed about
conditional probabilities and we have seen that how the conditional
probabilities can be calculated especially of using the Bayes’ Rule.
Bayes’ Rule is a very important theorem on conditional probability and
it helps us to solve the complex problem on probability theory. With
this now we have come to an end on our discussion on probability
theory.
Thank you!
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