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, . . . En 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, . . . En 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, . . . En , 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.
Events which are both mutually exclusive and exhaustive
This can be explained in 3 steps:
1. Let E1, E2, E3, . . . En 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, . . . En are both mutually exclusive and exhaustive.
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.
Link to a few more solved examples is given below:
In the next section, we will see Axiomatic approach to probability.
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