Set Theoretic Principles
Let us introduce a few notations, which would be
If ‘A’ and ‘B’ be any two events of the sample space
- 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
P(AÈB) = P(A) + P(B).
As in this case no sample point would
be present in AÇB.
q P(A¢) = 1 - P(A)
AÈA¢ = S and A and A¢ are mutually exclusive
P(AÈA¢) = P(A) + P(A¢) = P(S) = 1
P(A¢) = 1 - P(A).
= P(A) - P(AÇB)
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) -
q P(A¢ÈB¢) = 1 - P(AÇB)
(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(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
= 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 A1,
A2 L , An are
¢n` events, then P(A1
È A2 L ÈAn)
= +- L
q P (A È B) ≥ max (P(A), P(B), P(A) + P(B) - 1)
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) £
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.
three students S1, S2 and S3 appear at a
competitive examination. The probability of S1 coming first is 3
times that of S2 and the probability of S2 coming first
is three times that of S3. Find the probability of each coming
first. Also find the probability that S1 or S2 comes
Let P1, P2 and P3 be
the probabilities of S1, S2 and S3 coming
P1 = 3P2 and P2 = 3P3 also, P1
+ P2 +P3 = 1 (as one of them has to come first)
3P2 + P2 + =
P2 = ,
P1 = ,
Finally probability of S1 or
S2 coming first = P1 + P2 =
pack of 52 cards two cards are drawn at random. Find the probability of the
Both cards are of spade.
card is of spade and one card is of diamond.
total number of ways in which 2 cards can be drawn = 52C2
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 E1, then n(E1) = Number of elements
in E1 = Number of ways in which 2 cards can be selected out of 13
cards of spade
13C2 = =
Probability of E1 = P(E1) = .
E2 be the event that one card is of spade and one is of diamond then
n(E2) = number of elements in E2 =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. .
= 13C1´13C1 = 13´13
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.
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) = 9C3 = =
Let E1 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
= number of ways of selecting 3 tickets out of six blank tickets
= 6C3 = =
probability of not winning the prize by A is
Since P(E1) + P()
=1 \ P(E1) = 1 – P()
Or P(E1) = 1 –
Now for B
n(S) = number of ways of selecting 1 ticket out of 3
Or n(S) = 3C1 = 3.
Let E2 be the event of winning the prize by
= number of ways of selecting one ticket out of 1 prize ticket =1C1
probability of winning the prize by B.
\ The ratio of
the probabilities of winning the prizes by A and B