In the previous section, we saw various types of events. In this section, we will see algebra of events.
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In an earlier chapter on sets, we saw various operations that can be performed on sets.
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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)
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In our present chapter, we saw that, events are sets. So we can apply those operations here also.
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We know that all events are subsets of S. So the set S can be considered as the universal set.
Complementary event
This can be explained in 6 steps:
1. Consider the experiment of tossing a coin three times.
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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.
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Then we can pick out three outcomes, which are (H,H,T), (H,T,H) and (T,H,H).
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We can write a set E using these outcomes:
E = {(H,H,T), (H,T,H), (T,H,H)}
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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.
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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.
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If it is an element of E, then we say that:
Event E has occurred.
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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.
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So we can write:
♦ Set E’
♦ is same as the
♦ Complement of set E.
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We already know the significance of the complement of a set. See fig.1.19 in section 1.8.
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From what we learned from those lessons, we can write: E’ = S - E
The event “A or B”
This can be explained in 3 steps:
1. Consider an experiment whose sample space is S.
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Let A and B be two events associated with the experiment.
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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.
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Since both A and B are subsets of S, the set A∪B will also be a subset of S
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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
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So, if the event A∪B occurs, we can write:
Either A or B has occurred.
The event “A and B”
This can be explained in 3 steps:
1. Consider an experiment whose sample space is S.
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Let A and B be two events associated with the experiment.
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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.
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Since both A and B are subsets of S, the set A∩B will also be a subset of S
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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
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So, if the event A∩B occurs, we can write:
Both A and B has occurred.
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Let us see an example. It can be written in 5 steps:
1. Consider the experiment of rolling a die two times.
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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.
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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).
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We can write a set A using these six outcomes:
A = {(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
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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.
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Then, from the 36 outcomes, we can pick out three, which are (5,6), (6,6) and (6,5).
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We can write a set B using these three outcomes:
B = {(5,6), (6,6), (6,5)}
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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.
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This is because,
♦ First throw has given 6.
♦ Sum of two numbers is atleast 11.
The event “A but not B”
This can be explained in 3 steps:
1. Consider an experiment whose sample space is S.
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Let A and B be two events associated with the experiment.
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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.
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Since both A and B are subsets of S, the set A-B will also be a subset of S
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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)
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So, if the event A-B occurs, we can write:
A has occurred. But B has not occurred.
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:
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For this experiment, we can easily write S, A and B.
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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}
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.
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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.
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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.
In the next section, we will see Exhaustive Events.
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