Compliment of E
If E is an event of sample space S and É is the event that E does not happen then
P(É)=1P(E)
É is called the compliment of E
n(E) + n(É) = n(S)
P(E) + P(É) = 1
Consider the example
A room has 3 sockets for bulbs. From a collection of 8 bulbs out of which 4 are defective , 3 bulbs are selected at random and put in the sockets. Find the probability that the room is lit
Solution
As 3 bulbs are selected from 8 bulbs
n(S) = ^{8}C_{3}== 56 ways
Let A be the event that the room is lit
This gives rise to 3 cases
 Either 1 bulb is nondefective
 Either 2 bulbs are nondefective
 Either 3 bulbs are nondefective
This involves tedious calculations and lengthy work. Hence consider the event Á whih implies that the room is dark
Á = ^{4}C_{3} == 4 ways
P(Á) = 4/56 = 1/14
so P(A) = 1 P(Á)= 1 1/14 = 13/14
Solved Examples
 From a pack of 52 cards, 3 cards are drawn at random. Find the probabilty that
a)at least 1 is a heart.
b)all are not hearts.
Solution
n(S) = ^{52}C_{3}
a)let A be the event that at least one is a heart
So Á will be the event that non is a heart
n(Á) = ^{39}C_{3}
P(Á)=^{39}C_{3} /^{52}C_{3}
P(Á)=39x38x37/52x51x50 = 703/1700
So the required probability is
P(A) = 1 P(Á).
P(A) = 1703/1700 = 997/1700
b) Let B be the event that all are not hearts.
So P(&Bacute;) will be the event that all are hearts.
n(&Bacute;) = ^{13}C_{3}
P(&Bacute;) = 13x12x11/52x51x50 = 11/850
So the required probability is
P(B) = 1  11/850 = 839/850
 A committee of 4 is to be formed from 10 boys and 1 girl at random. Find the probability that the girl is included.
Solution
n(S) = ^{11}C_{4}
Let A be the event that the girl is included.
So Á will be the event that the girl is never included.
n(Á)= ^{10}C_{4}
P(Á)= 10x9x8x7 / 11x10x9x8 = 7/11
So the required probability is
P(A) = 1  7/11 = 4/11
