**● Complement**

(noun) something that completes an event; it adds what is missing to make up the whole.

**● Complementary**

(adjective) an event that completes or adds to other events to make up the whole sample space.

NOTE: A complement is not the same as a

__compliment__!

A

__compliment__is a positive comment made to a person or a group of people.

**● Complementary events**

Events that are

**mutually exclusive and make up the whole sample space**are called

**complementary events**. There is no intersection and no elements from the sample set are outside the two sets.

**● Complement of an Event**

Let

*A*be an event in a sample space

*S*the complement of

*A*is the set of all sample points of the space other than the sample point in

*A*and it is denoted by,

*A*‘ or

*Ā*∈

*S*,∉

*A*

(i).

*A*∪

*A*‘=

*S*

(ii).

*A*∩

*A*‘=Ø

The possible events when you roll a die are 1; 2; 3; 4; 5 or 6.

The probability of rolling a 4 is ⅙.

The probability of __not__ rolling a 4 is ⅚.

So the event not rolling a 4 is the complement of the event rolling a 4.

So P(4)+P(4′)=⅙+⅚=1

**● The complementary rule**

*A*‘)+P(

*A*)=1 or P(

*A*‘)=1-P(

*A*)

P(

*A*‘) means probability of ‘not

*A*‘.

In the example, n(not rolling a 4)+n(rolling a 4)=5+1=6

*S*: Possible outcomes for rolling a die

**● Complementary Event**

In throwing of a die consider a particular event

*E*={1,3,5}={getting an odd number on the face}

The complimentary event of

*E*, represented by

*Ē*={2,4,6}=not happening of event

*E*.

*)=1-P(*

*Ē**E*)or P(

*E*)+P(

*)=1*

*Ē***● Complementary Events**

The probability of complementary events refers to the probability associated with events not occurring. For example, if P(

*A*)=0.25, then the probability of

*A*not occurring is the probability associated with all other events in

*S*occurring less the probability of

*A*occurring. This means that

*)=1-P(*

*A*‘*A*)

where

*A*‘ or

*Ā*refers to ‘not

*A*‘ In other words, the probability of ‘not

*A*‘ is equal to one minus the probability of

*A*.

Q1. Jane invested in the stock market. The probability that she will not lose all her money is 0.32. What is the probability that she will lose all her money? Explain.

solution:

Since the events ”Jane will lose all her money” and ”Jane will not lose all her money” are complementary events, their probabilities sum to 1. Therefore the probability that Jane will lose all her money is 1-0.32=0.68.

Q2. A packet has yellow sweets and pink sweets. The probability of taking out a pink sweet is 7/12. What is the probability of taking out a yellow sweet?

solution:

a) wins the prize

solution:

b) does not win the prize

solution:

a) an Opel

solution:

190/300=19/30

b) not an Opel

solution:

a) orange

solution:

b) not orange

solution:

c) pink

solution:

d) not pink

solution:

e) orange or pink

solution:

f) neither orange nor pink

solution:

(i) Probability of an event

*E*+Probability of the event ‘not

*E*‘ =___.

(ii) The probability of an event that cannot happen is ___. Such as event is called ___.

(iii) The probability of an event that is certain to happen is ___. Such as event is called ___.

(iv) The sum of the probabilities of all the elementary events of an experiment is ___.

(v) The probability of an event is greater than or equal to ___ and less than or equal to ___.

solution:

(i) 1

(ii) 0, impossible event

(iii) 1, sure event or certain event

(iv) 1

(v) 0.1

Q7. If P(*E*)=0.59; find n(not *E*)

solution:

*E*)+P(not

*E*)=1

0.59+P(not

*E*)=1

P(not

*E*)=1-0.59=0.41

solution:

0.897+P(have same birthday)=1

P(have same birthday)=1-0.897=0.103

*E*)=0.05, what is the probability of ‘not

*E*‘?

solution:

We know that,

*)=1-P(*

*Ē**E*)

P(

*)=1-0.05=0.95*

*Ē*Therefore, the probability of ‘not

*E*‘ is 0.95.

Q10. A bag contains 6 red, 8 black and 4 white balls. A ball is drawn at random. What is the probability that ball drawn is not black?

Sol:

Total number of possible outcomes n(S)=18 {6 red, 8 black, 4 white}

Let *E*→ event of drawing black ball.

Number of favorable outcomes n(*E*)=8 {8 black balls}

*Ē*→ event of not drawing black ball

Q11. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?

solution:

(i) Total number of balls in the bag =8

(ii) Probability of getting a non-red ball =1- Probability of getting a red ball

Q12. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nun i will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that

(i) She will buy it?

(ii) She will not buy it?

solution:

Total number of pens =144

Total number of defective pens =20

Total number of good pens =144-20=124

(i) Probability of getting a good pen =124/144=31/36

(ii) P(Nuri will not buy a pen)=1-31/36=5/36

**● Types of events**

Events can be classified into various types on the basis of the elements they have.

1. Impossible and Sure Events. The empty set Ø and the sample space

*S*describe events. In fact Ø is called an

__impossible event__and

*S*, i.e., the whole sample space is called the

__sure event__.

To understand these let us consider the experiment of rolling a die. The associated sample space is

*S*={1,2,3,4,5,6}

Let

*E*be the event “

__the number appears on the die is a multiple of 7__”. Can you write the subset associated with the event

*E*?

Clearly no outcome satisfies the condition given in the event, i.e., no element of the sample space ensures the occurrence of the event

*E*. Thus, we say that the empty set only correspond to the event

*E*. In other words we can say that it is impossible to have a multiple of 7 on the upper face of the die. Thus, the event

*E*=Ø is an impossible event.

Now let us take up another event

*F*“the number turns up is odd or even”. Clearly,

*F*={1, 2, 3, 4, 5, 6,}=

*S*, i.e., all outcomes of the experiment ensure the occurrence of the event

*F*. Thus, the event

*F*=

*S*is a sure event.

2. Simple Event. If an event

*E*has only one sample point of a sample space, it is called a simple (or elementary) event.

In a sample space containing

*n*distinct elements, there are exactly

*n*simple events.

For example in the experiment of tossing two coins, a sample space is

*S*={HH,HT,TH,TT}

There are four simple events corresponding to this sample space. These are

*E*

_{1}={HH},

*E*

_{2}={HT},

*E*

_{3}={TH} and

*E*

_{4}={TT}.

3. Compound Event. If an event has more than one sample point, it is called a

__Compound event__.

For example, in the experiment of “tossing a coin thrice” the events

*E*: ‘Exactly one head appeared’

*F*: ‘

*A*tleast one head appeared’

*G*: ‘

*A*tmost one head appeared’ etc.

are all compound events. The subsets of S associated with these events are

*E*={HTT,THT,TTH}

*F*={HTT,THT,TTH,HHT,HTH,THH,HHH}

*G*={TTT,THT,HTT,TTH}

Each of the above subsets contain more than one sample point, hence they are all compound events.

**● Algebra of events**

In the Chapter on Sets, we have studied about different ways of combining two or more sets, viz, union, intersection, difference, complement of a set etc. Like-wise we can combine two or more events by using the analogous set notations.

Let

*A,B,C*be events associated with an experiment whose sample space is

*S*.

1. Complementary Event. For every event

*A*, there corresponds another event

*Ᾱ*called the complementary event to

*A*. It is also called the event ‘not

*A*‘.

For example, take the experiment ‘

__of tossing three coins__‘. An associated sample space is

*S*={HHH, HHT, HTH, THH, HTT, THT, TTH, ITT}

Let

*A*={HTH, HHT, THH} be the event ‘

__only one tail appears__‘.

Clearly for the outcome HTT, the event

*A*has not occurred. But we may say that the event ‘not

*A*‘ has occurred. Thus, with every outcome which is not in

*A*, we say that ‘not

*A*‘ occurs.

Thus the complementary event ‘not

*A*‘ to the event

*A*is

*Ᾱ*={HHH,HTT,THT,TTH,TTT}

or

*Ᾱ*={ω∶ω ϵ

*S*and ω∉

*A*}=

*S-A*

2. The Event ‘

*A*or

*B*‘. Recall that union of two sets

*A*and

*B*denoted by

*A*∪

*B*contains all those elements which are either in

*A*or in

*B*or in both.

When the sets

*A*and

*B*are two events associated with a sample space, then ‘

*A*∪

*B*‘ is the event ‘either

*A*or

*B*or both’. This event ‘

*A*∪

*B*‘ is also called ‘

*A*or

*B*‘.

Therefore Event ‘

*A*or

*B*‘=

*A*∪

*B*={ω∶ω ϵ

*A*or ω∶ω ϵ

*B*}.

3. The Event ‘

*A*and

*B*‘. We know that intersection of two sets

*A*∩

*B*is the set of those elements which are common to both

*A*and

*B*. i.e., which belong to both ‘

*A*and

*B*‘.

If

*A*and

*B*are two events, then the set

*A*∩

*B*denotes the event ‘

*A*and

*B*‘.

Thus,

*A*∩

*B*={ω∶ω ϵ

*A*and ω ϵ

*B*}

For example, in the experiment of ‘throwing a die twice’ Let

*A*be the event ‘score on the first throw is six’ and

*B*is the event ‘sum of two scores is atleast 11′ then

*A*={(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) } and

*B*={(5,6),(6,5),(6,6) }

so

*A*∩

*B*={(6,5),(6,6) }

Note that the set

*A*∩

*B*={(6,5),(6,6)} may represent the event ‘the score on the first throw is six and the sum of the scores is atleast 11′.

4. The Event ‘

*A*but not

*B*‘. We know that

*A*–

*B*is the set of all those elements which are in

*A*but not in

*B*. Therefore, the set

*A*–

*B*may denote the event ‘

*A*but not

B’. We know that

*A*–

*B*=

*A*∩

*B̄*

Example 6: Consider the experiment of rolling a die. Let

*A*be the event ‘getting a prime number’,

*B*be the event ‘getting an odd number’. Write the sets representing the events (i).

*A*or

*B*(ii).

*A*and

*B*(iii).

*A*but not

*B*(iv). ‘not

*A*‘.

solution: Here

*S*={1,2,3,4,5,6};

*A*={2,3,5} and B={1,3,5}

Obviously

(i). ‘

*A*or

*B*‘ =

*A*∪

*B*={1,2,3,5}

(ii). ‘

*A*and

*B*‘ =

*A*∩

*B*={3,5}

(iii). ‘

*A*but not

*B*‘ =

*A*–

*B*={2}

(iv). ‘not

*A*‘ =

*Ā*={1,4,6}

**● Probability of event ‘not**

*A*‘Consider the event

*A*={2,4,6,8} associated with the experiment of drawing a card from a deck of ten cards numbered from 1 to 10. Clearly the sample space is

*S*={1,2,3,…,10}

If all the outcomes 1,2,…,10 are considered to be equally likely, then the probabilityof each outcome is 1/10.

Also event ‘not

*A*‘=

*Ᾱ*={1,3,5,7,9,10}

*Ᾱ*)=P(1)+P(3)+P(5)+P(7)+P(9)+P(10)

=6/10=⅗

Thus, P(

*Ᾱ*)=⅗=1-⅖=1-P(

*A*)

Also, we know that

*Ᾱ*and

*A*are mutually exclusive and exhaustive events i.e,

*A*∩

*Ᾱ*=Ø and

*A*∪

*Ᾱ=S*or P(

*A*∪Ᾱ)=P(

*S*). Thus

*A*)+P(

*Ᾱ*)=1→P(not

*A*)=1-P(

*A*)

We now consider some examples and exercises having equally likely outcomes unless stated otherwise.

Q13. A bag contains three red balls, five white balls, two green balls and four blue balls:

1. Calculate the probability that a red ball will be drawn from the bag.

2. Calculate the probability that a ball which is not red will be drawn

answer:

Step 1: Find event 1

Let R be the event that a red ball is drawn:

●

*R*and

*R*‘ are complementary events

Step 2: Find the probability

Step 3 : Alternate way to solve it

● Alternately P(

*R*‘)=P(

*B*)+P(

*W*)+P(

*G*)