Let us introduce a few notations, which would be frequently used:

If ‘A’ and ‘B’ be any two events of the sample space then

- AÈB would stand for occurrence of atleast one of them.
- A Ç B stands for simultaneous occurrence of A and B.
- (or A¢) stands for non-occurrence of A
- (or A¢ Ç B¢) stands for non-occurrence of both A and B.
- A Í B stands for ‘the occurrence of B implies the occurrence of A’.

If A and B are any two events, then P(AÈB) = P(A) + P(B) - P(AÇB)

Consider the Venn – diagram in the adjacent figure. The P (AÈB) is the sum of weights of sample point in AÈB. Now P(A) + P(B) is the sum of all the weights in A plus the sum of all weights in B. Therefore we have added the weights in AÇB twice. Since these weights add up to give P(AÇB) we must subtract this probability once to obtain the sum of the weights in AÈB, which is P(A È B).

q If A and B are mutually exclusive,

P(AÈB) = P(A) + P(B).

As in this case no sample point would be present in AÇB.

q P(A¢) = 1 - P(A)

As, AÈA¢ = S and A and A¢ are mutually exclusive

&⇒
P(AÈA¢) = P(A) + P(A¢) = P(S) = 1

&⇒
P(A¢) = 1 - P(A).

q P(AÇB¢) = P(A) - P(AÇB)

As AÇB¢ and AÇB are mutually exclusive events and (AÇB¢) È (AÇB) = A

&⇒
P(AÇB¢) + P(AÇB) = P(A)

&⇒
P(AÇB¢) = P(A) - P(AÇB)

Similarly, P(A¢ÇB) = P(B) - P(AÇB)

q P(A¢ÈB¢) = 1 - P(AÇB)

As, (A¢ È B¢) È (A Ç B) = S

&⇒
P(A¢ È B¢)
+ P(A Ç B) = 1

&⇒
P(A¢ È B¢)
= 1 - P(AÇB)

Similarly, P(A¢ Ç B¢) = 1 - P(AÈB)

q P(exactly one of A, B occurs)

= P(AÇB¢) + P(A¢ÇB)

= P(A) + P(B) - 2P(AÇB)

= P(AÈB) - P(AÇB)

= P(A¢ È B¢) - P(A¢ Ç B¢)

If A, B, C are any three events of the sample space, then

q P(AÈBÈC) = P(A) + P(B) + P(C) - P(A Ç B) - P(AÇC) - P(BÇC) + P(AÇBÇC)

q P (Exactly one of A, B, C occurs)

= P(A) + P(B) + P(C) - 2P(AÇB) - 2P(AÇC) - 2P(BÇC) + 3P(AÇBÇC)

q P(Exactly two of A, B, C occur)

= P(AÇB) + P(BÇC) + P(AÇC) - 3P(AÇBÇC)

q P (at least two of A, B, C occur)

= P(AÇB) + P(BÇC) + P(AÇC) - 2P(AÇBÇC)

q If A_{1},
A_{2} L , A_{n} are
¢n` events, then P(A_{1}
È A_{2} L ÈA_{n})

= +- L +

q P (A È B) ≥ max (P(A), P(B), P(A) + P(B) - 1)

as
A Í AÈB

&⇒
P(A) £ P(AÈB)

Similarly, B Í A È B

&⇒
P(B) £ P(AÈB)

&⇒
P(AÈB) ≥ Max (P(A), P(B))

Also P(AÈB) = P(A) + P(B) - P(A Ç B)

and P(A) + P(B) - 1 £ P(AÈB) £ P(A) + P(B) (As 0 £ P(A Ç B) £ 1)

Max(P(A) + P(B) - 1, P(A), P(B)) £ P(A È B) £ P(A) + P(B)

q If out of m + n equally likely, mutually exclusive and exhaustive cases, m cases are favorable and n are not favorable to an event A, m : n is called odds in favour of A, n : m is called odds against the event A.

Only
three students S_{1}, S_{2} and S_{3} appear at a
competitive examination. The probability of S_{1} coming first is 3
times that of S_{2} and the probability of S_{2} coming first
is three times that of S_{3}. Find the probability of each coming
first. Also find the probability that S_{1 }or S_{2} comes
first.

Let P_{1}, P_{2} and P_{3} be
the probabilities of S_{1}, S_{2} and S_{3} coming
first respectively.

&⇒
P_{1} = 3P_{2} and P_{2} = 3P_{3} also, P_{1}
+ P_{2} +P_{3} = 1 (as one of them has to come first)

&⇒
3P_{2} + P_{2} + =
1

&⇒
P_{2} = ,
P_{1} = ,
P_{3 }=

Finally probability of S_{1 }or
S_{2} coming first = P_{1} + P_{2} =

From a pack of 52 cards two cards are drawn at random. Find the probability of the following events:.

(a) Both cards are of spade.

(b) One card is of spade and one card is of diamond.

The
total number of ways in which 2 cards can be drawn = ^{52}C_{2}

== 26´51 = 1326.

\ Number of elements in the space S are n(S) = 1326

(a) Let the event, that both cards are of
spade, be denoted by E_{1}, then n(E_{1}) = Number of elements
in E_{1} = Number of ways in which 2 cards can be selected out of 13
cards of spade

=
^{13}C_{2 }= =
78.

\
Probability of E_{1} = P(E_{1}) = .

(b) Let
E_{2} be the event that one card is of spade and one is of diamond then
n(E_{2}) = number of elements in E_{2} =number of ways in which
one card of spade can be selected out of 13 spade cards and one card of
diamond can be selected out of 13 diamond cards. .

= ^{13}C_{1}´^{13}C_{1} = 13´13
=169

\ P(E_{2})
= .

A has 3 shares in a lottery containing 3 prizes and 6 blanks; B has one share in a lottery containing 1 prize and 2 blanks. Compare their chances of success.

Total number of tickets in the first lottery = 3 + 6 =9

A may select any three tickets out of these 9 tickets. Therefore, the number of elements in the sample space S is given by .

= n(S) = ^{9}C_{3} = =
84

Let E_{1} be the event of winning the prize in
the first lottery by A. So, is
the event of not winning the prize of A, Then number of elements in is
given by

n() = number of ways of selecting 3 tickets out of six blank tickets

= ^{6}C_{3} = =
20

\The
probability of not winning the prize by A is

P()
=

Since P(E_{1}) + P()
=1 \ P(E_{1}) = 1 – P()

Or P(E_{1}) = 1 –

Now for B

n(S) = number of ways of selecting 1 ticket out of 3 tickets. .

Or n(S) = ^{3}C_{1} = 3.

Let E_{2} be the event of winning the prize by
B.

\ n(E_{2})
= number of ways of selecting one ticket out of 1 prize ticket =^{1}C_{1}
=1

\ The probability of winning the prize by B.

P(E_{2}) =

\ The ratio of the probabilities of winning the prizes by A and B

=.