×

Close

Type:
**Note**Course:
**
Placement Preparation
**Specialization:
**Quantitative Aptitude**Offline Downloads:
**2**Views:
**66**Uploaded:
**2 months ago**

PROBABILITY
Chapter
531
13
PROBABILITY
The theory of probabilities is simply the Science of logic
quantitatively treated. – C.S. PEIRCE
13.1 Introduction
In earlier Classes, we have studied the probability as a
measure of uncertainty of events in a random experiment.
We discussed the axiomatic approach formulated by
Russian Mathematician, A.N. Kolmogorov (1903-1987)
and treated probability as a function of outcomes of the
experiment. We have also established equivalence between
the axiomatic theory and the classical theory of probability
in case of equally likely outcomes. On the basis of this
relationship, we obtained probabilities of events associated
with discrete sample spaces. We have also studied the
addition rule of probability. In this chapter, we shall discuss
the important concept of conditional probability of an event
given that another event has occurred, which will be helpful
in understanding the Bayes' theorem, multiplication rule of
Pierre de Fermat
probability and independence of events. We shall also learn
(1601-1665)
an important concept of random variable and its probability
distribution and also the mean and variance of a probability distribution. In the last
section of the chapter, we shall study an important discrete probability distribution
called Binomial distribution. Throughout this chapter, we shall take up the experiments
having equally likely outcomes, unless stated otherwise.
13.2 Conditional Probability
Uptill now in probability, we have discussed the methods of finding the probability of
events. 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}

532
MATHEMATICS
1
to each sample point. Let
8
E be the event ‘at least two heads appear’ and F be the event ‘first coin shows tail’.
Then
E = {HHH, HHT, HTH, THH}
and
F = {THH, THT, TTH, TTT}
Therefore
P(E) = P ({HHH}) + P ({HHT}) + P ({HTH}) + P ({THH})
Since the coins are fair, we can assign the probability
1 1 1 1 1
+ + + = (Why ?)
8 8 8 8 2
P(F) = P ({THH}) + P ({THT}) + P ({TTH}) + P ({TTT})
=
and
1 1 1 1 1
+ + + =
8 8 8 8 2
E ∩ F = {THH}
=
Also
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.
with
P(E ∩ F) = P({THH}) =
Thus, Probability of E considering F as the sample space =
1
,
4
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).
or
Probability of E given that the event F has occurred =
1
4
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
P(E|F) =

PROBABILITY
533
Thus, we can also write the conditional probability of E given that F has occurred as
P(E|F) =
=
Number of elementary events favourable to E ∩ F
Number of elementary events which arefavourable to F
n (E ∩ F)
n (F)
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
n(E ∩ F)
P(E ∩ F)
n(S)
=
P(E|F) =
n(F)
P(F)
n(S)
... (1)
Note that (1) is valid only when P(F) ≠ 0 i.e., F ≠ φ (Why?)
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
P(E|F) =
P (E ∩ F)
provided P(F) ≠ 0
P (F)
13.2.1 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
P (S|F) =
P (S ∩ F) P (F)
=
=1
P (F)
P (F)
P(F|F) =
P (F ∩ F) P (F)
=
=1
P (F)
P (F)
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((A ∪ B)|F) = P (A|F) + P(B|F) – P ((A ∩ B)|F)

534
MATHEMATICS
In particular, if A and B are disjoint events, then
P((A∪B)|F) = P(A|F) + P(B|F)
We have
P((A ∪ B)|F) =
=
P[(A ∪ B) ∩ F]
P (F)
P[(A ∩ F) ∪ (B ∩ F)]
P (F)
(by distributive law of union of sets over intersection)
=
P (A ∩ F) + P (B ∩ F) – P (A ∩ B ∩ F)
P (F)
=
P (A ∩ F) P (B ∩ F) P[(A ∩ B) ∩ F]
+
−
P(F)
P(F)
P(F)
= P (A|F) + P (B|F) – P ((A ∩ B)|F)
When A and B are disjoint events, then
P ((A ∩ B)|F) = 0
⇒
P ((A ∪ B)|F) = P (A|F) + P (B|F)
Property 3 P (E′|F) = 1 − P (E|F)
From Property 1, we know that P (S|F) = 1
⇒
P (E ∪ E′|F) = 1
since S = E ∪ E′
⇒
P (E|F) + P (E′|F) = 1
since E and E′ are disjoint events
Thus,
P (E′|F) = 1 − P (E|F)
Let us now take up some examples.
Example 1 If P (A) =
7
9
4
, P (B) =
and P (A ∩ B) =
, evaluate P (A|B).
13
13
13
4
P (A ∩ B) 13 4
Solution We have P (A|B) =
=
=
9 9
P ( B)
13
Example 2 A family has two children. What is the probability that both the children are
boys given that at least one of them is a boy ?

## Leave your Comments