16.3 - Exhaustive Events

In the previous section, we saw algebra of events. We also saw mutually exclusive events. In this section, we will see Exhaustive events.

Exhaustive events

This can be explained in 7 steps:
1. Consider the experiment of rolling a die.
• We know that S = {1, 2, 3, 4, 5, 6}
2. Suppose that event A occurs if the number obtained is less than 4.
• Then we have: A = {1, 2,3}
3. Suppose that event B occurs if the number obtained is greater than 2 but less than 5.
• Then we have: B = {3,4}
4. Suppose that event C occurs if the number obtained is greater than 4.
• Then we have: C = {5,6}
5. Let us find A∪B∪C:
{1,2,3}∪{3,4}∪{5,6} = {1,2,3,4,5,6} = S
• We see that:
The union is same as S
6. Now we can write the definition of exhaustive events. It can be written in 2 steps:
(i) Let E₁, E₂, E₃, . . . E_n be some events associated with an experiment.
(ii) If the union of those events give the S of that experiment, then those events are called exhaustive events.
7. Let us see an interesting fact about exhaustive events. It can be written in 3 steps:
(i) Let E₁, E₂, E₃, . . . E_n be the exhaustive events associated with an experiment.
• The union of those events will be S. We can write:
$\rm{E_1 \cup E_2 \cup E_3 \cup~.~.~.~ \cup E_n~=~\cup_{i=1}^{i=n}{E_i}~=~S}$
(ii) We know that:
• When a set is formed by the union of two or more sets,
    ♦ All elements of all participating sets
    ♦ Will be present in the resulting set.
• So all elements of E₁, E₂, E₃, . . . E_n , will be present in $\rm{\cup_{i=1}^{i=n}{E_i}}$.
(iii) We also know that:
Whenever we perform the experiment, the outcome will be a member of S.
• But $\rm{S\,=\,\cup_{i=1}^{i=n}{E_i}}$
• So we can write:
Whenever the experiment is performed, one of the exhaustive events will surely occur.

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Events which are both mutually exclusive and exhaustive

This can be explained in 3 steps:
1. Let E₁, E₂, E₃, . . . E_n be some exhaustive events associated with an experiment.
2. Take any two sets from that list of exhaustive events. Those two sets must be disjoint.
• That is., which ever pair we take, the sets in that pair must be disjoint.
3. If the condition in (2) is satisfied, then we can say:
E₁, E₂, E₃, . . . E_n are both mutually exclusive and exhaustive.

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Now we will see some solved examples
Solved example 16.7
Two dice are thrown and the sum of the numbers which come up on the dice is noted. Let us consider the following events associated with the experiment.
A: the sum is even
B: the sum is a multiple of 3
C: the sum is less than 4
D: the sum is greater than 11
Which pairs of these events are mutually exclusive?
Solution:
1. We know that, the sample space is:
S ={
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
2. Now we can write the events:
(i) A: the sum is even
So A ={
(1,1), (1,3), (1,5),
(2,2), (2,4), (2,6),
(3,1), (3,3), (3,5),
(4,2), (4,4), (4,6),
(5,1), (5,3), (5,5),
(6,2), (6,4), (6,6)}
(ii) B: the sum is a multiple of 3
So B ={
(1,2), (1,5),
(2,1), (2,4),
(3,3), (3,6),
(4,2), (4,5),
(5,1), (5,4),
(6,3), (6,6)}
(iii) C: the sum is less than 4
So C ={
(1,1), (1,2), (2,1)}
(iv) D: the sum is greater than 11
So D = {(6,6)}
3. We have four sets: A, B, C and D
We must consider all possible pairs.
(i) Consider A and B
They are not disjoint sets because, there are some common elements. For example, (1,5).
(ii) Consider A and C
They are not disjoint sets because, there is one common element which is (1,1).
(iii) Consider A and D
They are not disjoint sets because, there is one common element which is (6,6).
(iv) Consider B and C
They are not disjoint sets because, there are two common elements: (1,2) and (2,1).
(v) Consider B and D
They are not disjoint sets because, there is one common element which is (6,6).
(vi) Consider C and D
They are disjoint sets because, there are no common elements.
4. So we can write:
Only one pair "C, D" are mutually exclusive events.

Solved example 16.8
A coin is tossed three times, consider the following events.
A: No head appears
B: Exactly one head appears
C: Atleast two heads appear
Do they form a set of mutually exclusive and exhaustive events?
Solution:
1. We know that, the sample space is:
S = {(H,H,H), (H,H,T), (H,T,H), (H,T,T), (T,H,H), (T,H,T), (T,T,H), (T,T,T)}
2. Now we can write the events:
(i) A: No head appears
So A = {(T,T,T)}
(ii) B: Exactly one head appears
So B = {(H,T,T), (T,H,T), (T,T,H)}
(iii) C: Atleast two heads appear
So C = {(H,H,H), (H,H,T), (H,T,H), (T,H,H)}
3. First we will check whether they are exhaustive events. For that, we must find A∪B∪C. We get:
{(T,T,T)}∪{(H,T,T), (T,H,T), (T,T,H)}∪ {(H,H,H), (H,H,T), (H,T,H), (T,H,H)}
= {(T,T,T), (H,T,T), (T,H,T), (T,T,H), (H,H,H), (H,H,T), (H,T,H), (T,H,H)}
• Compare this union with
S = {(H,H,H), (H,H,T), (H,T,H), (H,T,T), (T,H,H), (T,H,T), (T,T,H), (T,T,T)}
• We see that, A∪B∪C = S
So A, B and C are exhaustive events.
4. Next we check whether A, B and C are mutually exclusive. For that, we must consider all possible pairs.
(i) Consider A and B
They are disjoint sets because, there are no common elements.
(ii) Consider A and C
They are disjoint sets because, there are no common elements.
(iii) Consider B and C
They are disjoint sets because, there are no common elements.
5. So we can write:
A, B and C are mutually exclusive and exhaustive events.

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Link to a few more solved examples is given below:

Exercise 16.2

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In the next section, we will see Axiomatic approach to probability.

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16.2 - Algebra of Events

In the previous section, we saw various types of events. In this section, we will see algebra of events.

• In an earlier chapter on sets, we saw various operations that can be performed on sets.
• We saw these topics:
   ♦ Union of sets
   ♦ Intersection of sets
   ♦ Difference of sets
   ♦ Complement of a set etc.,
(see fig.1.5 in section 1.5)
• In our present chapter, we saw that, events are sets. So we can apply those operations here also.
• We know that all events are subsets of S. So the set S can be considered as the universal set.

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Complementary event

This can be explained in 6 steps:
1. Consider the experiment of tossing a coin three times.
• We know that S = {(H,H,H), (H,H,T), (H,T,H), (H,T,T), (T,H,H), (T,H,T), (T,T,H), (T,T,T)}
(see first example in exercise 16.1 )
2. Suppose that, we are interested in those outcomes in which:
   ♦ Exactly one T is obtained.
• Then we can pick out three outcomes, which are (H,H,T), (H,T,H) and (T,H,H).
• We can write a set E using these outcomes:
E = {(H,H,T), (H,T,H), (T,H,H)}
• So set E corresponds to the event in which T is obtained only once.
3. We picked out three out of the eight outcomes in S. So there are five outcomes remaining.
• If any one of those five outcomes occur, we can say that:
Event E has not occurred.
4. When an outcome occurs, we check whether it is an element of E.
• If it is an element of E, then we say that:
Event E has occurred.
• If it is not an element of E, then we say that:
Event "not E" has occurred.
5. If E is an event, then "not E" is also an event.
• "not E" is also known as the complementary event to E.
   ♦ It is denoted as E’.
6. Clearly, the set corresponding to "not E" will contain all elements of S except the elements of E.
• So we can write:
   ♦ Set E’
   ♦ is same as the
   ♦ Complement of set E.
• We already know the significance of the complement of a set. See fig.1.19 in section 1.8.
• From what we learned from those lessons, we can write: E’ = S - E

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The event “A or B”

This can be explained in 3 steps:
1. Consider an experiment whose sample space is S.
• Let A and B be two events associated with the experiment.
• Then both A and B will be subsets of S
2. Since both A and B are sets, we can write a new set: A∪B.
• Since both A and B are subsets of S, the set A∪B will also be a subset of S
• Since A∪B is a subset of S, we can write:
A∪B is an event.
(Recall that, all subsets of S are events)
3. A∪B will contain elements of both A and B
• So, if the event A∪B occurs, we can write:
Either A or B has occurred.

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The event “A and B”

This can be explained in 3 steps:
1. Consider an experiment whose sample space is S.
• Let A and B be two events associated with the experiment.
• Then both A and B will be subsets of S
2. Since both A and B are sets, we can write a new set: A∩B.
• Since both A and B are subsets of S, the set A∩B will also be a subset of S
• Since A∩B is a subset of S, we can write:
A∩B is an event.
(Recall that, all subsets of S are events)
3. A∩B will contain only those elements which are present in both A and B
• So, if the event A∩B occurs, we can write:
Both A and B has occurred.

• Let us see an example. It can be written in 5 steps:
1. Consider the experiment of rolling a die two times.
• We know the 36 elements of S
(see second example in exercise 16.1 )
2. Suppose that, we are interested in those outcomes in which:
   ♦ Number in the first throw is 6.
• Then from the 36 outcomes, we can pick out six, which are (6,1), (6,2), (6,3), (6,4), (6,5) and (6,6).
• We can write a set A using these six outcomes:
A = {(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
• So set A corresponds to the event in which 6 is obtained in the first throw.
3. Also suppose that, we are interested in those outcomes in which:
   ♦ Sum of the numbers in the two throws is atleast 11.
• Then, from the 36 outcomes, we can pick out three, which are (5,6), (6,6) and (6,5).
• We can write a set B using these three outcomes:
B = {(5,6), (6,6), (6,5)}
• So set B corresponds to the event in which sum of the numbers in the two throws is atleast 11.
4. Now we can find A∩B.
We get: A∩B = {(6,5), (6,6)}
5. When the experiment gives an outcome which is an element of A∩B, we can write:
Both A and B has occurred.
• This is because,
   ♦ First throw has given 6.
   ♦ Sum of two numbers is atleast 11.

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The event “A but not B”

This can be explained in 3 steps:
1. Consider an experiment whose sample space is S.
• Let A and B be two events associated with the experiment.
• Then both A and B will be subsets of S
2. Since both A and B are sets, we can write a new set: A-B.
• Since both A and B are subsets of S, the set A-B will also be a subset of S
• Since A-B is a subset of S, we can write:
A-B is an event.
(Recall that, all subsets of S are events)
3. A-B will contain only those elements which are present in A but not in B.
(see fig.1.16 in section 1.7)
• So, if the event A-B occurs, we can write:
A has occurred. But B has not occurred.

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Now we will see a solved example
Solved example 16.6
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:
• For this experiment, we can easily write S, A and B.
• We have:
   ♦ S = {1, 2, 3, 4, 5, 6}
   ♦ A = {2, 3, 5}
   ♦ B = {1, 3, 5}
Part (i): A or B
1. If the event A∪B occurs, we can say A or B has occurred.
2. We have: A∪B = {1, 2, 3, 5}
Part (ii): A and B
1. If the event A∩B occurs, we can say A and B has occurred.
2. We have: A∩B = {3, 5}
Part (iii): A but not B
1. If the event A-B occurs, we can say A has occurred, but  B has not occurred.
2. We have: A-B = {2}
Part (iv): "not A"
1. If the event S-A occurs, we can say "not A" has occurred.
2. We have: S-A = {1, 4, 6}

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Mutually exclusive events

This can be explained in 4 steps:
1. We have seen the case of "A and B"
• If an outcome in A∩B occurs, we can say: Both A and B has occurred.
• It is clear that there is atleast one element in A∩B.
2. What if there are no elements in A∩B? (That is.,A∩B = Φ)
• This happens when there are no elements common to A and B. In other words, A and B are disjoint sets.
• In such a situation, we will be able to write two points:
(i) If A occurs, B has not occurred.
(ii) If B occurs, A has not occurred.
3. Two events A and B are called mutually exclusive events if occurrence of any one of them excludes the occurrence of the other event.
• In other words, mutually exclusive events cannot occur simultaneously.
4. We have seen that, if S have n elements, then there will be n simple events.
• Those n simple events are mutually exclusive events. When any one of them occurs, we can readily say that the remaining (n-1) events have not occurred. This is because, each of the n simple events have only one element. They cannot have any common elements.

• Let us see an example for mutually exclusive events. It can be written in 4 steps:
1. Consider the experiment of rolling a die.
• We know that S = {1, 2, 3, 4, 5, 6}
2. Suppose that event A occurs if the number obtained is odd.
Then we have: A = {1, 3, 5}
3. Suppose that event B occurs if the number obtained is even.
Then we have: B = {2, 4, 6}
4. We see that, A and B are disjoint sets. A and B will never occur simultaneously. 

• Let us see an example for two events which are "not mutually exclusive". Such an example will help us to get a better understanding about events which are actually "mutually exclusive". It can be written in 5 steps:
1. Consider the experiment of rolling a die.
• We know that S = {1, 2, 3, 4, 5, 6}
2. Suppose that event A occurs if the number obtained is odd.
Then we have: A = {1, 3, 5}
3. Suppose that event B occurs if the number obtained is less than 4.
• Then we have: B = {1, 2, 3}
4. We see that, A and B are not disjoint sets.
• We can write:
    ♦ If 1 is obtained, A and B has occurred simultaneously.
    ♦ If 3 is obtained, A and B has occurred simultaneously.
5. So in this case, A and B are not mutually exclusive events.

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In the next section, we will see Exhaustive Events.

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Chapter 16.1 - Event in Probability

In the previous section, we completed a discussion on sample space. In this section, we will see event.

Event can be explained in 9 steps:
1. Consider the experiment of tossing a coin twice.
• We know that the sample space is:
S = {(H,H), (H,T), (T,H), (T,T)}
2. Suppose that, we are interested in those outcomes in which H occurs exactly once.
• Then we can pick out two outcomes: (H,T) and (T,H).
• We can form a set E using the two outcomes that we picked out:
E = {(H,T), (T,H)}
• We can write 3 points:
(i) We are interested in those outcomes in which H occurs exactly once.
• So we say that:
In the experiment, if H occurs exactly once, we have an event.
(ii) Two outcomes are favorable for the event.
• We write a set E in such a way that, those two outcomes are the only elements.
(iii) Then E will be a subset of S.
3. Suppose that, we are interested in those outcomes in which T occurs exactly two times.
• Then we can pick out one outcome: (T,T).
• We can form a set E using the outcome that we picked out:
E = {(T,T)}
• We can write 3 points:
(i) We are interested in those outcomes in which T occurs exactly two times.
• So we say that:
In the experiment, if T occurs exactly two times, we have an event.
(ii) One outcome is favorable for the event.
• We write a set E in such a way that, that one outcome is the only element.
(iii) Then E will be a subset of S.
4. Suppose that, we are interested in those outcomes in which T occurs at least once.
• Then we can pick out three outcomes: (H,T), (T,H) and (T,T).
• We can form a set E using the three outcomes that we picked out:
E = {(H,T), (T,H), (T,T)}
• We can write 3 points:
(i) We are interested in those outcomes in which T occurs at least once.
• So we say that:
In the experiment, if T occurs at least once, we have an event.
(ii) Three outcomes are favorable for the event.
• We write a set E in such a way that, those three outcomes are the only elements.
(iii) Then E will be a subset of S.
5. Suppose that, we are interested in those outcomes in which number of H is atmost 1.
("atmost 1" means, the number must not exceed 1. In other words, maximum allowed is 1)
• Then we can pick out three outcomes: (H,T), (T,H) and (T,T).
• We can form a set E using the three outcomes that we picked out:
E = {(H,T), (T,H), (T,T)}
• We can write 3 points:
(i) We are interested in those outcomes in which number of H is atmost 1.
• So we say that:
In the experiment, if H occurs atmost one time, we have an event.
(ii) Three outcomes are favorable for the event.
• We write a set E in such a way that, those three outcomes are the only elements.
(iii) Then E will be a subset of S.
6. Suppose that, we are interested in those outcomes in which second toss is not head.
• Then we can pick out two outcomes: (H,T) and (T,T).
• We can form a set E using the two outcomes that we picked out:
E = {(H,T), (T,T)}
• We can write 3 points:
(i) We are interested in those outcomes in which second toss is not H.
• So we say that:
In the experiment, if second toss is not H, we have an event.
(ii) Two outcomes are favorable for the event.
• We write a set E in such a way that, those two outcomes are the only elements.
(iii) Then E will be a subset of S.
7. Suppose that, we are interested in those outcomes in which number of T is atmost 2.
• Then we can pick out all the four outcomes: (H,H), (H,T), (T,H) and (T,T).
• We can form a set E using the four outcomes that we picked out:
E = {(H,H), (H,T), (T,H), (T,T)}
• We can write 3 points:
(i) We are interested in those outcomes in which number of T is atmost 2.
• So we say that:
In the experiment, if number of T is atmost 2, we have an event.
(ii) Four outcomes are favorable for the event.
• We write a set E in such a way that, those four outcomes are the only elements.
(iii) Then E will be a subset of S.
8. Suppose that, we are interested in those outcomes in which number of T is more than 2.
• Then we can pick out none of the four outcomes.
• We can form only a null set. A null set is also a subset of S.
• We can write 3 points:
(i) We are interested in those outcomes in which number of T is more than 2.
• So we say that:
In the experiment, if number of T is more than 2, we have an event.
(ii) No outcome is favorable for the event.
• We write a set E which is a null set.
(iii) Then E is a subset of S.
(Recall that, null set is also a subset)
9. Based on the above steps, we can write a definition for event. It can be written in 2 steps:
(i) An event is a set. It is denoted using the letter ‘E’
(ii) It is a subset of S.
• So all elements of E are outcomes.

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Occurrence of an event

This can be explained in 4 steps:
1. Consider the experiment of rolling a die.
• We know that S = {1, 2, 3, 4, 5, 6}
2. We are interested in those outcomes which are less than 4.
• Then we can pick out three outcomes: 1, 2 and 3.
• We can write a set E using the three outcomes that we picked out:
E = {1, 2, 3}
3. Now we can write about the occurrence of the event:
• When the die is rolled, if 1 is obtained, then we say:
Event E has occurred.
• When the die is rolled, if 2 is obtained, then we say:
Event E has occurred.
• When the die is rolled, if 3 is obtained, then we say:
Event E has occurred.
4. Based on the above three steps, we can write the definition for “occurrence of event”. It can be written in 3 steps:
(i) Let an outcome 𝛚 of an experiment occur.
(ii) Let 𝛚 be an element of E. In other words, 𝛚 ∈ E.
• Then we say that:
Event E has occurred.
(iii) If 𝛚 ∉ E, then we say that:
Event E has not occurred.

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Types of events

There are four types of events.
(i) Impossible event  (ii) Sure event  (iii) Simple event  (iv) Compound event.

Impossible event

This can be explained in 3 steps:
1. Consider the experiment of rolling a die.
• We know that S = {1, 2, 3, 4, 5, 6}
2. Suppose that, we are interested in those outcomes which are multiples of 7.
• Then we can pick out no outcomes.
• So the set E will be a null set.
3. If E is a null set, then that event is an impossible event.
• We can do the experiment any number of times we like. We will never get an outcome which is an element of E. Because, E is a null set.

Sure event

This can be explained in 3 steps:
 1. Consider the experiment of rolling a die.
• We know that S = {1, 2, 3, 4, 5, 6}
2. Suppose that, we are interested in those outcomes which are either odd or even.
• Then we can pick out all six outcomes.
• We can write a set E using those outcomes:
E = {1, 2, 3, 4, 5, 6}
   ♦ We see that, E is same as S.
3. If E = S, then that event is a sure event.
• We can do the experiment any number of times we like. The outcome will always be an element of E. Because, all outcomes are present in E.

Simple event

This can be explained in 7 steps:
1. Consider the experiment of tossing a coin two times.
We know that S = {(H,H), (H,T), (T,H), (T,T)}
2. Suppose that, we are interested in those outcomes in which:
   ♦ First toss gives T.
   ♦ Second toss gives H.
• Then we can pick out only one outcome, which is (T,H).
• We can write a set E using this outcome:
E = {(T,H)}
3. If E has only one element, then that event is called a simple event.
4. In fact, we can pick out each element from S and write distinct sets.
   ♦ E₁ = {(H,H)}
   ♦ E₂ = {(H,T)}
   ♦ E₃ = {(T,T)}
• E₁ is the event in which both tosses give H.
• E₂ is the event in which first toss gives H and second toss gives T.
• E₃ is the event in which both tosses give T.
5. So we can write an important point:
If there are n elements in S, then there will be n simple events.
6. Consider the event in which there is atleast one H.
• This event is not a simple event because, the set of this event has more than one elements.
7. A simple event is also known as an elementary event.    

Compound event

This can be explained in 5 steps:
1. Consider the experiment of tossing a coin three times.
We know that S = {(H,H,H), (H,H,T), (H,T,H), (H,T,T), (T,H,H), (T,H,T), (T,T,H), (T,T,T)}
(see first example in exercise 16.1 of the previous section)
2. Suppose that, we are interested in those outcomes in which:
   ♦ Exactly one H is obtained.
• Then we can pick out three outcomes, which are (H,T,T), (T,H,T) and (T,T,H).
• We can write a set E using these outcomes:
E = {(H,T,T), (T,H,T), (T,T,H)}
3. If E has more than one element, then that event is called a compound event.
4. We can write more examples from this experiment. Let us write two such examples:
Example (i):
Suppose that, we are interested in those outcomes in which:
   ♦ Atleast one H is obtained.
• Then we can write:
   ♦ E₁ = {(H,H,H), (H,H,T), (H,T,H), (H,T,T), (T,H,H), (T,H,T), (T,T,H), (T,T,T)}
• E₁ is a compound event because there are more than one elements.
Example (ii):
Suppose that, we are interested in those outcomes in which:
   ♦ Atmost one H is obtained.
• Then we can write:
   ♦ E₂ = {(H,T,T), (T,H,T), (T,T,H), (T,T,T)}
• E₂ is a compound event because there are more than one elements.
5. So we can write an important point:
All subsets of S, which have more than one elements, are compound events.

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In the next section, we will see Algebra of Events.

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