**Conditional probability**

Uptill now in probability, we have discussed the methods of finding the probability of events we will see now conditional probability of an event.

If we have two events from the same sample space, does the information about the occurrence of one of the events affect the probability of the other event? Let us try to answer this question by taking up a random experiment in which the outcomes are equally likely to occur.

Consider the experiment of tossing three fair coins. The sample space of the experiment is

S = { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Since the coins are fair, we can assign the probability 1/ 8 to each sample point. Let E be the event ‘at least two heads appear’ and F be the event ‘first coin shows tail’.

Then E={HHH, HHT, HTH, THH}

F={THH, THT, TTH, TTT}

Therefore P(E) = P({HHH}) +P({HHT}) +P({HTH}) +P({THH})

and P(F) = P({THH}) + P({THT}) +P({TTH}) + P({TTT})

Also E∩F = {THH}

with P(E∩F) = P({THH}) = 1/8

Now, suppose we are given that the first coin shows tail, i.e. F occurs, then what is the probability of occurrence of E? With the information of occurrence of F, we are sure that the cases in which first coin does not result into a tail should not be considered while finding the probability of E. This information reduces our sample space from the set S to its subset F for the event E. In other words, the additional information really amounts to telling us that the situation may be considered as being that of a new random experiment for which the sample space consists of all those outcomes only which are favourable to the occurrence of the event F.

Now, the sample point of F which is favourable to event E is THH.

Thus, Probability of E considering F as the sample space = 1 /4,

or Probability of E given that the event F has occurred = 1 /4,

This probability of the event E is called the conditional probability of E given that F has already occurred, and is denoted by P (E|F).

Thus P(E/F) =

Note that the elements of F which favour the event E are the common elements of E and F, i.e. the sample points of E ∩ F.

Thus, we can also write the conditional probability of E given that F has occurred as

Dividing the numerator and the denominator by total number of elementary events of the sample space, we see that P(E|F) can also be written as

Note that (1) is valid only when P(F) ≠ 0 i.e., F ≠ φ

Thus, we can define the conditional probability as follows :

Definition 1 If E and F are two events associated with the same sample space of a random experiment, the conditional probability of the event E given that F has occurred,

i.e. P (E|F) is given by

**Properties of conditional probability**

Let E and F be events of a sample space S of an experiment, then we have** Property 1** –P (S|F) = P(F|F) = 1

We know that

Also

Thus

P(S/F) = P(F/F) =1

**Property 2** -If A and B are any two events of a sample space S and F is an event of S such that P(F) ≠ 0, then P((AUB)/F) = P(A/F) +P(B/F) -P((A∩B)/F)

In particular, if A and B are disjoint events, then

P((A∪B)|F) = P(A|F) + P(B|F)

We have

When A and B are disjoint events, then

Property 3 P(E′|F) = 1 − P (E|F)

From Property 1, we know that P (S|F) = 1

**Example** If

Solution We have

**Example-** In a school, there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in Class XII given that the chosen student is a girl?** Solution-** Let E denote the event that a student chosen randomly studies in Class XII and F be the event that the randomly chosen student is a girl. We have to find P (E|F).

Now

Then