Consider the following situations:
i. two events occur successively or one after the other
ii. both event A and event B occur together.

**Conditional
Probability**

Consider the following situations:

i. two events occur successively or one after the other (e.g) A
occurs after B has occurred and

ii. both event *A* and
event *B* occur together.

There are 4000 people living in a village including 1500 female.
Among the people in the village, the age of 1000 people is above 25 years which
includes 400 female. Suppose a person is chosen and you are told that the
chosen person is a female. What is the probability that her age is above 25
years?

Here, the event of interest is selecting a female with age above
25 years. In connection with the occurrence of this event, the following two
events must happen.

A:a person selected is female

B:a person chosen is above 25 years.

We are interested in the event *B*, given that *A* has
occurred. This event can be denoted by *B*|*A*. It can be read as “*B* given *A*”. It means that first the event A occurs then under that condition,
B occurs. Here, we want to find the probability for the occurrence of *B*|*A*
i.e., *P*(*B*|*A*). This probability is
called conditional probability. In reverse, the probability for selecting a
female given that a person has been selected with age above 25 years is denoted
by *P*(*A*|*B*).

Suppose that it is interested to select a person who is both
female and with age above 25 years. This event can be denoted by A ∩B.

Calculation of probabilities in these situations warrant us to
have another theorem namely Multiplication theorem. It is derived based on the
definition of conditional probability.

If *P*(*B*) > 0, the conditional probability
of *A* given *B* is defined as

If P(B) = 0, then P(A∩B) = 0. Hence, the above formula is
meaningless when P(B) = 0. Therefore, the conditional probability P(A|B) can be
calculated only when P(B)>0.

The need for the computation of conditional probability is
described in the following illustration.

**Illustration**

A family is selected at random from the set of all families in a
town with one twin pair. The sample space is

S = {(boy, boy), (boy, girl), (girl, boy), (girl, girl)}.

Define the events

*A*: the randomly
selected family has two boys, and

*B*: the randomly selected
family has a boy.

Let us assume that all the families with one twin pair are
equally likely. Since

A = {(boy, boy)},

B = {(boy, boy), (boy, girl), (girl, boy)},

*A**∩**B *= A = {(boy, boy)}.

Applying the classical definition of probability, it can be
calculated that

Suppose that the randomly selected family has a boy. Then, the
probability that the other child in the pair is a girl can be calculated using
conditional probability as

A number is selected randomly from the digits11 through 19.
Consider the events

*A *= { 11,14, 16, 18, 19
}

*B *= { 12, 14, 18, 19 }

*C *= { 13, 15, 18, 19 }.

Find (i) P(A/B) (ii) P(A/C) (iii) P(B/C) (iv) P (B/A)

Therefore, the probability for the occurrence of A given that B
has occurred is

The probability for the occurrence of *A* given that *C* has
occurred is

Similarly, the conditional probability of B given C is

and the conditional probability of *B* given *A* is

A pair of dice is rolled and the faces are noted. Let

*A*: sum of the faces is
odd,* B*: sum of the faces exceeds 8,
and

*C*: the faces are
different then find (i) P (*A*/*C*) (ii) *P* (*B*/*C*)

The outcomes favourable to the occurrence of these events are

A = { (1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4),
(3,6), (4,1), (4,3), (4,5), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5) }

B = { (3,6), (4,5),
(4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6) }

C = { (1,2), (1,3),
(1,4), (1,5), (1,6), (2,1), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,4),
(3,5), (3,6), (4,1), (4,2), (4,3), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,6), (6,1),
(6,2), (6,3), (6,4), (6,5) }

Since A and B are proper
subsets of C, A∩C = A and B∩C = B.

Hence,
the probability for the sum of the faces is an odd number given that the faces
are different is

Similarly,
the probability for the sum of the faces exceeds 8 given that the faces are
different is

The
conditional probabilities also satisfy the same axioms introduced in Section
8.3.

If
*S* is the sample space of a random experiment and B is an
event in the experiment, then** **

(i)
P(A/B) ≥ 0 for any event A of S.

(ii)
P(S/B) = 1

(iii)
If A_{1}, A_{2}, … is a sequence of mutually exclusive events,
then

In
continuation of conditional probability, another property of events, viz.,
independence can be studied. It is discussed in the next section. Also,
multiplication theorem, a consequence of conditional probability, will be
studied later.

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11th Statistics : Chapter 8 : Elementary Probability Theory : Conditional Probability |

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