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High School Mathematics - 2
13.4 Compliment of an Event

Compliment of E


If E is an event of sample space S and is the event that E does not happen then
P()=1-P(E)
is called the compliment of E
n(E) + n() = n(S)
P(E) + P() = 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) = 8C3== 56 ways
Let A be the event that the room is lit
This gives rise to 3 cases
  • Either 1 bulb is non-defective
  • Either 2 bulbs are non-defective
  • Either 3 bulbs are non-defective

This involves tedious calculations and lengthy work. Hence consider the event whih implies that the room is dark
= 4C3 == 4 ways
P() = 4/56 = 1/14
so P(A) = 1- P()= 1- 1/14 = 13/14

Solved Examples

  1. 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) = 52C3
    a)let A be the event that at least one is a heart
    So will be the event that non is a heart
    n() = 39C3
    P()=39C3 /52C3
    P()=39x38x37/52x51x50 = 703/1700
    So the required probability is
    P(A) = 1- P().
    P(A) = 1-703/1700 = 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;) = 13C3
    P(&Bacute;) = 13x12x11/52x51x50 = 11/850
    So the required probability is
    P(B) = 1 - 11/850 = 839/850

  2. 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) = 11C4
    Let A be the event that the girl is included.
    So will be the event that the girl is never included.
    n()= 10C4
    P()= 10x9x8x7 / 11x10x9x8 = 7/11 So the required probability is
    P(A) = 1 - 7/11 = 4/11

Q 1: 8 Indians and 3 Americans are to stand in a row at random. Find the probability that no 2 Americans are always together
24/55
28/55
19/55
27/55

Q 2: 2 students, Julia and Kate are given a problem to solve. The chance that Julia solves the problem is 1/5 and that of Kate is 1/6. What is the probability that the problem is solved.
3/5
1/3
1/6
2/7

Q 3: The probabilities of 3 people A,B,C winning a lottery is 0.5, 0.2, 0.3. Find the probabiltiy that the lottery is not won
0.49
0.45
0.89
0.28

Q 4: The probabilities that 3 students Mike, John , Linda will solve a problem in Algebra is 0.2, 0.7, 0.6 respectively . Find the probability that the problem will not be solved
0.009
0.906
0.609
0.096

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