Event
We have studied about random experiment and sample space associated with an experiment. The sample space serves as an universal set for all questions concerned with the experiment.
Consider the experiment of tossing a coin two times. An associated sample space is \(S =\{ HH , HT , TH , TT \}\).
Now suppose that we are interested in those outcomes which correspond to the occurrence of exactly one head. We find that \(HT \text { and } TH\) are the only elements of \(S\) corresponding to the occurrence of this happening (event). These two elements form the set \(E=\{ HT , TH \}\)
We know that the set \(E\) is a subset of the sample space \(S\). Similarly, we find the following correspondence Eventbetween events and sub
sets of \(S\).
\(
\begin{array}{ll}
\text { Description of events } & \text { Corresponding subset of ‘S’ } \\
\text { Number of tails is exactly } 2 & A=\{ TT \} \\
\text { Number of tails is atleast one } & B =\{ HT , TH , TT \} \\
\text { Number of heads is atmost one } & C =\{ HT , TH , TT \} \\
\text { Second toss is not head } & D =\{ HT , TT \} \\
\text { Number of tails is atmost two } & S =\{ HH , HT , TH , TT \} \\
\text { Number of tails is more than two } & \phi
\end{array}
\)
The above discussion suggests that a subset of sample space is associated with an event and an event is associated with a subset of sample space. In the light of this we define an event as follows.
Definition: Any subset \(E\) of a sample space \(S\) is called an event.
Occurrence of an event
Consider the experiment of throwing a die. Let E denotes the event ” a number less than 4 appears”. If actually ‘ 1 ‘ had appeared on the die then we say that event \(E\) has occurred. As a matter of fact if outcomes are 2 or 3 , we say that event E has occurred
Thus, the event E of a sample space S is said to have occurred if the outcome \(\omega\) of the experiment is such that \(\omega \in E\). If the outcome \(\omega\) is such that \(\omega \notin E\), we say that the event E has not occurred.
Types of events
Events can be classified into various types on the basis of the elements they have.
Case-I: Impossible and Sure Events
The empty set \(\phi\) and the sample space \(S\) describe events. In fact \(\phi\) 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 \(E =\phi\) 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.
Case-II: 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
\(
E _1=\{ HH \}, E _2=\{ HT \}, E _3=\{ TH \} \text { and } E _4=\{ TT \} \text {. }
\)
Case III: 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’
\(F\): ‘Atleast one head appeared’
\(G\): ‘Atmost one head appeared’ etc.
are all compound events. The subsets of S associated with these events are
\(
\begin{aligned}
& E =\{\text { HTT,THT,TTH }\} \\
& F =\{\text { HTT,THT, TTH, HHT, HTH, THH, HHH }\} \\
& G =\{\text { TTT, THT, HTT, TTH }\}
\end{aligned}
\)
Each of the above subsets contain more than one sample point, hence they are all compound events.
Case-IV: Equally likely Events
The given events are said to be equally likely, if none of them is expected to occur in preference to the other.
For example,
Case-V: Independent Events
Two events are said to be independent, if the occurrence of one does not depend on the occurrence of the other.
For example, When an unbiased die is thrown, then the sample space \(S=\{1,2,3,4,5,6\}\)
Let \(E_1=\{1,3,5\}=\) the event of occurrence of an odd number and \(E_2=\{2,4,6\}=\) the event of occurrence of an even number. Clearly, the occurrence of odd number does not depend on the occurrence of even number. So, \(E_1\) and \(E_2\) are independent events.
Case-VI: Complementary Event
Let \(E\) be an event and \(S\) be the sample space for a random experiment, then complement of \(E\) is denoted by \(E^{\prime}\) or \(E^c\) or \(\bar{E}\). Clearly, \(E^{\prime}\) means \(E\) does not occur.
Thus, \(E^{\prime}\) occurs \(\Leftrightarrow E\) does not occur.
For example, When an unbiased die is thrown, then the sample space \(S=\{1,2,3,4,5,6\}\).
If
\(
E=\{1,4,6\} \text {, then } E^{\prime}=\{2,3,5\}
\)
Case-VII: Mutually Exclusive Events
A set of events is said to be mutually exclusive, if occurrence of one of them precludes the occurrence of any of the remaining events. If a set of events \(E_1, E_2, \ldots, E_n\) for mutually exclusive events.
Then,
\(
E_1 \cap E_2 \cap \ldots \cap E_n=\phi
\)
For example, If we thrown an unbiased die, then the sample space \(S=\{1,2,3,4,5,6\}\) in which \(E_1=\{1,2,3\}=\) the event of occurrence of a number less than 4 and \(E_2=\{5,6\}=\) the event of occurrence of a number greater than 4. Clearly, \(E_1 \cap E_2=\phi\) So, \(E_1\) and \(E_2\) are mutually exclusive.
Case-VIII: Exhaustive Events
A set of events is said to be exhaustive, if the performance of the experiment results in the occurrence of atleast one of them. If a set of events \(E_1, E_2, \ldots, E_n\) for exhaustive events.
Then, \(\quad E_1 \cup E_2 \cup \ldots \cup E_n=S\)
For example, If we thrown an unbiased die, then sample space \(S=\{1,2,3,4,5,6\}\) in which
\(E_1=\{1,2,3,4\}=\) the event of occurrence of a number less than 5 and \(E_2=\{3,4,5,6\}=\) the event of occurrence of a number greater than 2 .
Then, \(E_1 \cup E_2=\{1,2,3,4,5,6\}\) and \(E_1 \cap E_2=\{3,4\}\)
So, \(\quad E_1 \cup E_2=S\) and \(E_1 \cap E_2 \neq \phi\)
Hence, \(E_1\) and \(E_2\) are exhaustive events.
Case-IX: Mutually Exclusive and Exhaustive Events
A set of events is said to be mutually exclusive and exhaustive, if above two conditions are satisfied. If a set of events \(E_1, E_2, \ldots, E_n\) for mutually exclusive and exhaustive events.
Then, \(E_1 \cup E_2 \cup \ldots \cup E_n=S\) and \(E_1 \cap E_2 \cap \ldots \cap E_n=\phi\) For example, If we thrown an unbiased die, then sample space
\(S=\{1,2,3,4,5,6\}\) in which
\(E_1=\{1,3,5\}=\) the event of occurrence of an odd number and \(E_2=\{2,4,6\}=\) the event of occurrence of an even number.
Then, \(E_1 \cup E_2=\{1,2,3,4,5,6\}\) and \(E_1 \cap E_2=\phi\)
So, \(E_1 \cup E_2=S\) and \(E_1 \cap E_2=\phi\).
Hence, \(E_1\) and \(E_2\) are mutually exclusive and exhaustive 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 \(S\).
Case-1: Complementary Event
For every event \(A\), there corresponds another event \(A ^{\prime}\) 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
\(
S =\{ HHH , HHT , HTH , THH , \text { HTT, THT, TTH, TTT }\}
\)
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
\(
A ^{\prime}=\{ HHH , HTT , THT , TTH , TTT \}
\)
\(
\text { or } \quad A ^{\prime}=\{\omega: \omega \in S \text { and } \omega \notin A \}= S – A \text {. }
\)
Case-2: \(\text { The Event ‘A or B’ }\)
Recall that union of two sets \(A\) and \(B\) denoted by \(A \cup B\) 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 \(\cup B\) ‘ is the event ‘either \(A\) or \(B\) or both’. This event ‘ \(A \cup B\) ‘ is also called ‘\(A\) or \(B\) ‘.
Therefore \(\quad\) Event ‘ A or \(B ^{\prime}= A \cup B\)
\(=\{\omega: \omega \in A\) or \(\omega \in B\}\)
Case-3: \(\text { The Event ‘A and B’ }\)
We know that intersection of two sets \(A \cap B\) 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 \(A \cap B\) denotes the event ‘ \(A\) and \(B\) ‘.
Thus, \(\quad A \cap B =\{\omega: \omega \in A\) and \(\omega \in B \}\)
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 \(A \cap B =\{(6,5),(6,6)\}\)
Note that the set \(A \cap B =\{(6,5),(6,6)\}\) may represent the event ‘the score on the first throw is six and the sum of the scores is atleast 11’.
Case-4: \(\text { The Event ‘A but not B’ }\)
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
\(
A – B = A \cap B ^{\prime}
\)
Example 1: Consider the experiment of rolling a die. Let \(\text { A }\) be the event ‘getting a prime number’, \(\text { B }\) 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:
\(
\text { Here } \quad S =\{1,2,3,4,5,6\}, A =\{2,3,5\} \text { and } B =\{1,3,5\}
\)
Obviously
\(
\text { (i) ‘A or } B ^{\prime}= A \cup B =\{1,2,3,5\}
\)
\(
\text { (ii) ‘ } A \text { and } B \text { ‘ }= A \cap B =\{3,5\}
\)
\(
\text { (iii) ‘A but not } B \text { ‘ }= A – B =\{2\}
\)
\(
\text { (iv) ‘not } A ^{\prime}= A ^{\prime}=\{1,4,6\}
\)
Case-5: Mutually exclusive events
In the experiment of rolling a die, a sample space is \(S =\{1,2,3,4,5,6\}\). Consider events, \(A\) ‘an odd number appears’ and \(B\) ‘an even number appears’
Clearly the event \(A\) excludes the event \(B\) and vice versa. In other words, there is no outcome which ensures the occurrence of events \(A\) and \(B\) simultaneously. Here
\(
A =\{1,3,5\} \text { and } B =\{2,4,6\}
\)
Clearly \(A \cap B =\phi\), i.e., \(A\) and \(B\) are disjoint sets.
In general, two events \(A\) and \(B\) are called mutually exclusive events if the occurrence of any one of them excludes the occurrence of the other event, i.e., if they can not occur simultaneously. In this case the sets \(A\) and \(B\) are disjoint.
Again in the experiment of rolling a die, consider the events \(A\) ‘an odd number appears’ and event \(B\) ‘a number less than 4 appears’
Obviously \(A =\{1,3,5\}\) and \(B =\{1,2,3\}\)
Now \(3 \in A\) as well as \(3 \in B\)
Therefore, \(A\) and \(B\) are not mutually exclusive events.
Remark: Simple events of a sample space are always mutually exclusive.
Case-6: Exhaustive events
Consider the experiment of throwing a die. We have \(S=\{1,2,3,4,5,6\}\). Let us define the following events
\(\text { A: }\) ‘a number less than 4 appears’,
\(\text { B: }\) ‘a number greater than 2 but less than 5 appears’ and
\(\text { C: }\) ‘a number greater than 4 appears’.
Then \(A =\{1,2,3\}, B =\{3,4\}\) and \(C =\{5,6\}\). We observe that \(A \cup B \cup C=\{1,2,3\} \cup\{3,4\} \cup\{5,6\}=S\).
Such events \(A , B\) and \(C\) are called exhaustive events. In general, if \(E _1, E _2, \ldots, E _{ n }\) are \(n\) events of a sample space \(S\) and if
\(
E _1 \cup E _2 \cup E _3 \cup \ldots \cup E _n=\cup_{i=1}^n E _i= S
\)
then \(E _1, E _2, \ldots ., E _{ n }\) are called exhaustive events. In other words, events \(E _1, E _2, \ldots, E _n\) are said to be exhaustive if atleast one of them necessarily occurs whenever the experiment is performed.
Further, if \(E _i \cap E _j=\phi\) for \(i \neq j\) i.e., events \(E _i\) and \(E _j\) are pairwise disjoint and \(\cup_{i=1}^n E _i= S\), then events \(E _1, E _2, \ldots, E _n\) are called mutually exclusive and exhaustive events.
Example 2: 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 this 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: There are 36 elements in the sample space \(S =\{(x, y)\) : \(x, y=1,2,3,4,5,6\}\).
Then
\(
\begin{aligned}
A = & \{(1,1),(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4), \\
& (4,6),(5,1),(5,3),(5,5),(6,2),(6,4),(6,6)\} \\
B = & \{(1,2),(2,1),(1,5),(5,1),(3,3),(2,4),(4,2),(3,6),(6,3),(4,5),(5,4), \\
& (6,6)\} \\
C = & \{(1,1),(2,1),(1,2)\} \text { and } D =\{(6,6)\}
\end{aligned}
\)
We find that
\(
A \cap B=\{(1,5),(2,4),(3,3),(4,2),(5,1),(6,6)\} \neq \phi
\)
Therefore, \(A\) and \(B\) are not mutually exclusive events.
Similarly \(A \cap C \neq \phi, A \cap D \neq \phi, B \cap C \neq \phi\) and \(B \cap D \neq \phi\).
Thus, the pairs of events, \((A, C), (A, D), (B, C), (B, D)\) are not mutually exclusive events.
Also \(C \cap D =\phi\) and so \(C\) and \(D\) are mutually exclusive events.
Example 3: \(A\) coin is tossed three times, consider the following events.
\(A\): ‘No head appears’, \(B\): ‘Exactly one head appears’ and \(C\): ‘Atleast two heads appear’.
Do they form a set of mutually exclusive and exhaustive events?
Solution: The sample space of the experiment is
\(
S =\{ HHH , HHT , HTH , THH , HTT , THT , TTH , TTT \}
\)
\(
\text { and } A =\{ TTT \}, B =\{ HTT , THT , TTH \}, C =\{ HHT , HTH , THH , HHH \}
\)
Now \(A \cup B \cup C =\{ TTT , HTT , THT , TTH , HHT , HTH , THH , HHH \}= S\)
Therefore, \(A , B \text { and } C\) are exhaustive events. Also, \(A \cap B =\phi, A \cap C =\phi\) and \(B \cap C =\phi\)
Therefore, the events are pair-wise disjoint, i.e., they are mutually exclusive. Hence, \(A , B \text { and } C\) form a set of mutually exclusive and exhaustive events.
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