1.Algebra Booster

Probability 8.3
A set of events E 1
, E 2
, …, E n
of a sample space S is said to be
mutually exclusive and exhaustive event if
(i) E 1
> E 2
> E 3
> … E n–1
> E n
= j
(ii) E 1
> E 2
> E 3
> … E n–1
> E n
= S.
To explain further, we use some examples.
EXAMPLE 1: If we throw an unbiased die, then S = {1, 2, …, 6}
in which E 1
= {2, 4, 6} and E 2
= {1, 3, 5}.
Clearly E 1
< E 2
= S and E 1
< E 2
= j.
Thus E 1
and E 2
are mutually exclusive and exhaustive
events.
EXAMPLE 2: Let a die is thrown. The sample is S = {1, 2, 3, 4,
5, 6}.
Let A = {2, 4, 6} and B = {1, 2, 3, 5}
Then A and B are not mutually exclusive, i.e. A « B π j
But A » B = {1, 2, 3, 4, 5, 6} = S.
Thus A and B are exhaustive events.
From the above examples, we can conclude that, mutually
exclusive events can be a exhaustive events and its converse
is also true.
Independent Events
Two or more events are said to be independent when the occurrence
of one does not affect the other.
For example, if we toss a coin twice, the occurrence of
the second toss will, in no way, be affected by the outcome
of the first toss.
But if we throw a die, the sample space is
S = {1, 2, …, 6}
Let E be the event of getting an even number and F be the
event of getting an odd number, then
E = {2, 4, 6} and F = {1, 3, 5} clearly E 1
> E 2
= j.
Thus E and F are mutually exclusive events. But these
events are not independent.
The event of getting a tail on the first coin and the event
of getting a tail on the second coin in a simultaneous throw of
two coins are independent.
Probability of Occurrence of an Event
Let S be a sample space and A be any event.
Then
nA ( )
P( A ) =
nS ( )
Number of cases
favourable to event A
=
total number of cases
If A be any event and A¢ be the complement event of A on
a sample S, then
S
P(A) + P(A¢) = 1
Proof
A¢
Here A and A¢ are mutually exclusive
and exhaustive event of a sample space
A
S.
S
A
Thus A « A¢ = j and A » A¢ = S
Now, P(A » A¢) = P(S) = 1
fi P(A) + P(A¢) = 1.
fi P(A) = 1 – P(A¢).
Odds in Favour and Against of an Event
Let S be a sample space and A be an event. Let A¢ be the
complement of an event A, then
(i) Odds in favour of an event A
number of casesfavourable to theevent A
=
number of casesagainst theevent A
nA ( ) nA ( )/ nS ( ) PA ( )
= = =
nA ( ¢ ) nA ( ¢ )/ nS ( ) PA ( ¢ )
(ii) Odds against an event A
number of casesagainst theevent A
=
number of casesfavourable to theevent A
n A¢ P A¢
= =
n A P A
( )
( )
( )
( )
5. AXIOMS OF PROBABILITY
Let A be any event of a sample space S. Then
Axioms 1: 0 £ P(A) £ 1
Axioms 2: P(S) = 1.
Axioms 3:
Ê ˆ
PÁ
A˜
= Â P( Ai)
Ë
i
i= 1 ¯ i=
1
6. ADDITION THEOREM ON PROBABILITY
Theorem: If A and B be any two events of a sample space
S, then
P(A » B) = P(A) + P(B) – P(A « B)
Notes
(i) If A and B be mutually exclusive events, then
P(A » B) = P(A) + P(B).
(ii) If A, B and C be any three events, then
P(A » B » C)
= P(A) + P(B) + P(C) – P(A « B)
–P(A « C) – P(B « C) + P(A « B « C)
(iii) If A, B and C be three mutually exclusive events, then
P(A » B » C) = P(A) + P(B) + P(C)
(iv) The probability of an event at least one of the events
A and B is P(A » B).
(v) P(A – B) = P(A) – P(A « B)
(vi) P(B – A) = P(B) – P(A « B)
7. INEQUALITIES IN PROBABILITY
Let S be a sample space and A and B be two events associated
with the same sample space.
(i) If A be a proper subset of B, i.e. A Ã B, then
P(A) £ P(B)