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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Chapter 16 - Probability

In the previous section, we completed a discussion on statistics. In this chapter, we will see probability.

In our earlier classes, we have seen the basics of probability. The links to those notes are given below:

    ♦ Probability part I consists of chapters 1.5, 1.6, . . . up to 1.9
    ♦ probability part II consists of chapters 28, 28.1, . . . up to 28.4
    ♦ probability part III consists of chapters 36, 37.1 and 36.2

The reader must have a thorough knowledge on the above three parts. In our present discussion, we will see some advanced techniques.

First we must become familiar with four basic terms:
(i) Random experiments, (ii) Outcomes  (iii) Sample space (iv) Sample point.

Random experiments

This can be explained in 4 steps:
1. Consider an experiment in which a student is analyzing the sum of interior angles of various triangles.
• He takes the first triangle and measure it’s three interior angles. He finds that the sum of those three angles is 180^o   
• He takes the second triangle and measure it’s three interior angles. He finds that the sum of those three angles is 180^o   
• He can take any number of triangles. The sum will be always 180^o.
• So the experiment of “analyzing sum of interior angles of triangles” will always give a fixed result. We can predict the result. We can predict that, the sum will be 180^o
2. Consider the experiment of tossing a coin.
• A student tosses a coin. The result can be head or tail. We cannot predict the result.
• The student tosses the coin again. This time also, the result can be head or tail. We cannot predict the result.
• In this experiment,
    ♦ There are two possible results: head and tail.
    ♦ It is not possible to predict the result.
• Such experiments are called random experiments.
• Note that, if there is only one possible result, a need for prediction will not arise.
• We can write:
An experiment is a random experiment if it satisfies two conditions:
(i) It has more than one possible result.
(ii) It is not possible to predict the result in advance.   
3. Based on the above steps, we can write:
The experiment of rolling a die is a random experiment.
• This is because, there are six possible results. Also, it is not possible to predict the result.
4. In this chapter, whenever we mention the word “experiment”, it will be a random experiment.

Outcomes  

This can be explained in 4 steps:
1. Consider the experiment of rolling a die.
2. The possible results are: 1, 2, 3, 4, 5 and 6
3. Each of the above possible results is called an outcome of the experiment.
4. So we can write:
    ♦ Outcomes of an experiment
    ♦ are
    ♦ possible results of that experiment.

Sample space

This can be explained in 4 steps:
1. Consider the experiment of rolling a die.
2. The outcomes are: 1, 2, 3, 4, 5 and 6
3. Using these outcomes, we can form a set S:
S = {1, 2, 3, 4, 5, 6}
• This set is called sample space of the experiment.
4. So we can define sample space in 3 steps:
(i) Sample space of an experiment, is a set
(ii) All outcomes of that experiment will be elements of that set.
(iii) This set is denoted by the letter ‘S’.

Sample point

This can be explained in 2 steps:
1. Each element of S is called a sample point.
2. But we know that, each element is an outcome.
• So we can write:
Each outcome is a sample point.

---

Let us see some solved examples:
Solved Example 16.1
Two different coins are tossed once. Find the sample space.
Solution:
1. Given that the coins are different.
    ♦ So the head of one coin will be different from the head of the other coin.
    ♦ Also, the tail of one coin will be different from the tail of the other coin.
• Because of this difference, we will give specific names:
    ♦ Head of first coin can be named as H1.
    ♦ Head of the other coin can be named as H2.
    ♦ Tail of first coin can be named as T1.
    ♦ Tail of the other coin can be named as H2.
2. Let us write the possible outcomes:
(i) Outcome 1:
    ♦ First coin gives H1.
    ♦ The other coin gives H2.
• Note that, in this outcome, T1 and T2 are not possible. For example, the first coin landed with H1 on the upper side. So T1 will not be visible.
• We can write this outcome as: (H1,H2)
(ii) Outcome 2:
    ♦ First coin gives H1.
    ♦ The other coin gives T2.
• Note that, in this outcome, T1 and H2 are not possible.
• We can write this outcome as: (H1,T2)
(iii) Outcome 3:
    ♦ First coin gives T1.
    ♦ The other coin gives H2.
• Note that, in this outcome, H1 and T2 are not possible.
• We can write this outcome as: (T1,H2)
(iv) Outcome 4:
    ♦ First coin gives T1.
    ♦ The other coin gives T2.
• Note that, in this outcome, H1 and H2 are not possible.
• We can write this outcome as: (T1,T2)
3. Based on the above outcomes, we can write the sample space as:
S = {(H1,H2), (H1,T2), (T1,H2), (T1,T2)}

Solved Example 16.2
Two different dice (one blue and the other red) are rolled once. Find the sample space.
Solution:
1. We have already seen this type of problem in our earlier classes (see solved example 1.14 in section 1.9).
2. In our present case, we are asked to write the sample space.
• For presenting the sample space in a “convenient and easy to understand” manner, we use ordered pairs (x,y)
• If we take the blue die as the first die, then:
    ♦ All values associated with blue die are x.
    ♦ All values associated with red die are y.
3. Based on such a notation, the sample space for our present case can be written as:
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)}
• There are 36 elements in S.

Solved example 16.3
In each of the following experiments, write the sample space.
(i) A boy has a 1 rupee coin, a 2 rupee coin and a 5 rupee coin in his pocket. He takes two coins out of his pocket, one after the other.
(ii) A person is noting down the number of accidents along a busy highway during a year.
Solution:
Part (i):
1. We will use the form of ordered pairs (x,y)
    ♦ x is related to the first coin taken out.
    ♦ y is related to the second coin taken out.
2. Let us write the outcomes:
(i) First coin is 1, second coin is 2.
We write this as (1,2).
(ii) First coin is 1, second coin is 5.
We write this as (1,5)    
(iii) First coin is 2, second coin is 1.
We write this as (2,1)    
(iv) First coin is 2, second coin is 5.
We write this as (2,5)    
(v) First coin is 5, second coin is 1.
We write this as (5,1)    
(vi) First coin is 5, second coin is 2.
We write this as (5,2)
3. So the sample space can be written as:
S = {(1,2), (1,5), (2,1), (2,5), (5,1), (5,2)}
• There are 6 elements in S.
Part (ii):
1. The number of accidents in a year may be 0 (if there are no accidents), or any positive integer.
2. So the sample space can be written as:
S = {0, 1, 2, 3, . . . }

Solved example 16.4
A coin is tossed. If it shows head, we draw a ball from a bag containing 3 blue and 4 white balls. If it shows tail, we through a die. Write the sample space of this experiment.
Solution:
1. We will use the form of ordered pairs (x,y)
    ♦ x is related to tossing of coin.
    ♦ y is related to drawing a ball or rolling the die.
2. Let us write the out comes:
(i) The coin shows heads. The ball drawn is the first blue ball.
• We write this as (H,B₁).    
(ii) The coin shows heads. The ball drawn is the second blue ball.
• We write this as (H,B₂).    
(iii) The coin shows heads. The ball drawn is the third blue ball.
• We write this as (H,B₃).
(iv) The coin shows heads. The ball drawn is the first white ball.
• We write this as (H,W₁).
(v) The coin shows heads. The ball drawn is the second white ball.
• We write this as (H,W₂).   
(vi) The coin shows heads. The ball drawn is the third white ball.
• We write this as (H,W₃).    
(vii) The coin shows heads. The ball drawn is the fourth white ball.
• We write this as (H,W₄).
(viii) The coin shows tails. The die shows 1.
We write this as (T,1).         
(ix) The coin shows tails. The die shows 2.
We write this as (T,2).         
(x) The coin shows tails. The die shows 3.
We write this as (T,3).         
(xi) The coin shows tails. The die shows 4.
We write this as (T,4).         
(xii) The coin shows tails. The die shows 5.
We write this as (T,5).         
(xiii) The coin shows tails. The die shows 6.
We write this as (T,6).
3. So the sample space can be written as:
S = {(H,B₁), (H,B₂), (H,B₃), (H,W₁), (H,W₂), (H,W₃), (H,W₄), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
• There are 13 elements in S.

Solved Example 16.5
Consider the experiment in which a coin is tossed repeatedly until a head comes up. Write the sample space.
Solution:
1. Let us write the outcomes:
(i) Head is obtained in the first toss.
We write this as H.
(ii) Head is obtained only in the second toss.
We write this as TH.
(iii) Head is obtained only in the third toss.
We write this as TTH.
(iv) Head is obtained only in the fourth toss.
We write this as TTTH.
so on . . .
2. So the sample space can be written as:
S = {H, TH, TTH, TTTH, . . . }

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Link to a few more solved examples is given below:

Exercise 16.1

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In the next section, we will see Event.

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Chapter 15.7 - Miscellaneous Examples

In the previous section, we completed a discussion on the coefficient of variance. In this section, we will see some miscellaneous examples.

Solved example 15.17
The variance of 20 observations is 5. If each observation is multiplied by 2, find the new variance of the resulting observations.
Solution:
1. Analysis of the original data set:
(i) Let the observations in the original data set be denoted as $x_i$.
• Then we can write: ${\sigma^2}_x~=~\frac{\sum(x_i - \bar x)^2}{20}~=~5$
(ii) In the above equation, we know that: $\bar x ~=~\frac{\sum x_i}{20}$
(iii) Expanding the equation in (i), we get:

$\begin{array}{ll}
{}&{5}
& {~=~}& {\frac{\sum(x^2_i - 2 x_i \bar x + {\bar x}^2)}{20}}
&{} \\

{\Rightarrow}&{5}
& {~=~}& {\frac{\sum{x^2_i} ~-~ 2 \bar x \sum{x_i} ~+~ \sum{{\bar x}^2}}{20}}
&{} \\

{\Rightarrow}&{100}
& {~=~}& {\sum{x^2_i} ~-~ 2 \bar x \sum{x_i} ~+~ \sum{{\bar x}^2}}
&{} \\

\end{array}$

2. Analysis of the modified data set:
(i) Let the observations in the modified data set be denoted as $y_i$.
• Then we can write: ${\sigma^2}_y~=~\frac{\sum(y_i - \bar y)^2}{20}~=~K$
   ♦ We are asked to find K.
(ii) In the above equation, we know that: $\bar y ~=~\frac{\sum y_i}{20}$
(iii) Expanding the equation in (i), we get:

$\begin{array}{ll}
{}&{K}
& {~=~}& {\frac{\sum(y^2_i - 2 y_i \bar y + {\bar y}^2)}{20}}
&{} \\

{\Rightarrow}&{K}
& {~=~}& {\frac{\sum{y^2_i} ~-~ 2 \bar y \sum{y_i} ~+~ \sum{{\bar y}^2}}{20}}
&{} \\

{\Rightarrow}&{20K}
& {~=~}& {\sum{y^2_i} ~-~ 2 \bar y \sum{y_i} ~+~ \sum{{\bar y}^2}}
&{} \\

\end{array}$

3. Writing y in terms of x:
(i) Given that, each observation in the original data set is multiplied by 2, to get the modified data set.
(ii) So we can write: $y_i~=~ 2 x_i$
• Substituting this in 2(iii), we get:

$\begin{array}{ll}
{}&{20K}
& {~=~}& {\sum{y^2_i} ~-~ 2 \bar y \sum{y_i} ~+~ \sum{{\bar y}^2}}
&{} \\

{\Rightarrow}&{20K}
& {~=~}& {\sum{(2 x_i)^2} ~-~ 2 \bar y \sum{2 x_i} ~+~ \sum{{\bar y}^2}}
&{} \\

{\Rightarrow}&{20K}
& {~=~}& {\sum{(2 x_i)^2} ~-~ 2 (2 \bar x) \sum{2 x_i} ~+~ \sum{(2 \bar x)^2}~~ \color {green} {\text{- - - (I)}}}
&{} \\

{\Rightarrow}&{20K}
& {~=~}& {4 \sum{x^2_i} ~-~ 8 \bar x \sum{x_i} ~+~ 4 \sum{{\bar x}^2}}
&{} \\

{\Rightarrow}&{20K}
& {~=~}& {4 \left[ \sum{x^2_i} ~-~ 2 \bar x \sum{x_i} ~+~ \sum{{\bar x}^2}\right]~~ \color {green} {\text{- - - (II)}}}
&{} \\

{\Rightarrow}&{20K}
& {~=~}& {4 \left[100 \right]}
&{} \\

{\Rightarrow}&{20K}
& {~=~}& {400}
&{} \\

{\Rightarrow}&{K}
& {~=~}& {20}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
In this line, we use an information that we learned in our earlier classes. It can be written in 3 steps:
(i) Average of a set of observations is 'A'
(ii) The set is modified by multiplying each observation by 'm'.
(iii) Then the average of the modified set is 'mA'.
• So in our present case, we get: $\bar y = 2 \bar x$

• Line marked as II:
Based on the result in 1(iii), the portion inside the square brackets is '100'.

4. Now, based on 2(i), we can write:
${\sigma^2}_y~=~K~=~20$
• Thus we get the new variance of the modified observations.
• Note that, 20 = 2² × 5
• So we can write:
If the observations are multiplied by a constant 'm', then the variance of the modified observations will be m² times the original variance.

Solved example 15.18
The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations.
Solution:
1. We can use the usual table that is used for calculation of variance. It is shown below:

Table 15.29

• In column I, we fill up the observations ($x_i$). The two unknown observations are written as x₄ and x₅.
• In column II, we fill up the frequencies. Each of the five observations occur only once. So all values in the column II are '1'.
• In column III, we write the product $f_i x_i$. Since all $f_i$ values are '1', Column III will be same as column II.
• In column IV, we write the deviations from the mean. We are already given the mean. It is 4.4.
• In column V, we write the squares of the deviations. Here $f_i$ do not have any effect because, all $f_i$ values are '1'.
• After filling all the columns, we can start the calculations.
2. We know that, the column III is used to find mean.
• Sum of all the values in column III is:
1 + 2 + 6 + x₄ + x₅ = 9 + x₄ + x₅
• So we can write:
Mean = $\frac{9 + x_4 + x_5}{5}~=~4.4$
• From this, we get:
x₄ + x₅ = 13
3. We know that, the column V is used to find the variance.
Sum of all values in column V can be calculated as follows:

$\begin{array}{ll}
{}&{\text{Sum}}
& {~=~}& {11.56 + 5.76 + 2.56 + (x_4 - 4.4)^2 + (x_5 - 4.4)^2} &{} \\

{}&{}
& {~=~}& {19.88 + (x_4 - 4.4)^2 + (x_5 - 4.4)^2} &{} \\

{}&{}
& {~=~}& {19.88 ~+~{x_4}^2 - 8.8 x_4 + 19.36~+~{x_5}^2 - 8.8 x_5 + 19.36} &{} \\

{}&{}
& {~=~}& {{x_4}^2 +{x_5}^2 - 8.8 (x_4 + x_5) + 19.36 + 19.36 + 19.88} &{} \\

{}&{}
& {~=~}& {{x_4}^2 +{x_5}^2 - 8.8 (x_4 + x_5) + 58.6} &{} \\

{}&{}
& {~=~}& {{x_4}^2 +{x_5}^2 - 8.8 (13) + 58.6~\color{green}{\text{- - - I}}} &{} \\

{}&{}
& {~=~}& {{x_4}^2 +{x_5}^2 - 114.4 + 58.6} &{} \\

{}&{}
& {~=~}& {{x_4}^2 +{x_5}^2 - 55.8} &{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
In this line, we use the result from (2) to replace the sum $(x_4 + x_5)$

4. Based on the above sum of the squares, we can write:

$\begin{array}{ll}
{}&{\text{Variance}}
& {~=~}& {\frac{{x_4}^2 +{x_5}^2 - 55.8}{5}}
&{} \\

{\Rightarrow}&{8.24}
& {~=~}& {\frac{{x_4}^2 +{x_5}^2 - 55.8}{5}}
&{} \\

{\Rightarrow}&{41.2}
& {~=~}& {{x_4}^2 +{x_5}^2 - 55.8}
&{} \\

{\Rightarrow}&{97}
& {~=~}& {{x_4}^2 +{x_5}^2}
&{} \\

\end{array}$

5. Now we have two useful results:
   ♦ From (2) we have: $x_4 + x_5 = 13$
   ♦ From (3) we have: ${x_4}^2 +{x_5}^2 = 97$
• Using these two results, we can calculate $2 x_4 x_5$. This is shown below:

$\begin{array}{ll}
{}&{(x_4 + x_5)^2}
& {~=~}& {13^2}
&{} \\

{\Rightarrow}&{{x_4}^2 +2 x_4 x_5 + {x_5}^2 }
& {~=~}& {169}
&{} \\

{\Rightarrow}&{97 +2 x_4 x_5}
& {~=~}& {169}
&{} \\

{\Rightarrow}&{2 x_4 x_5}
& {~=~}& {169 - 97}
&{} \\

{\Rightarrow}&{2 x_4 x_5}
& {~=~}& {72}
&{} \\

\end{array}$

6. Consider the two results:
   ♦ From (3) we have: ${x_4}^2 +{x_5}^2 = 97$
   ♦ From (4) we have: $2 x_4 x_5 = 72$
• Using these two results, we can calculate $ x_4 - x_5$. This is shown below:

$\begin{array}{ll}
{}&{(x_4 - x_5)^2}
& {~=~}& {{x_4}^2 -2 x_4 x_5 + {x_5}^2 }
&{} \\

{\Rightarrow}&{(x_4 - x_5)^2}
& {~=~}& {{x_4}^2 + {x_5}^2 -2 x_4 x_5}
&{} \\

{\Rightarrow}&{(x_4 - x_5)^2}
& {~=~}& {97 - 72}
&{} \\

{\Rightarrow}&{(x_4 - x_5)^2}
& {~=~}& {25}
&{} \\

{\Rightarrow}&{x_4 - x_5}
& {~=~}& {\pm 5}
&{} \\

\end{array}$

7.  Consider the three equations:
(i) From (2) we have: $x_4 + x_5 = 13$
(ii) From (5) we have: $x_4 - x_5 = 5$
(iii) From (5) we have: $x_4 - x_5 = -5$
• Solving (i) and (ii), we get:
x₄ = 9, x₅ = 4
Solving (i) and (iii), we get:
• x₄ = 4, x₅ = 9
8. So the unknown observations are: 4 and 9

Solved example 15.19
Show that adding or subtracting a constant number 'a' from each observation, does not affect the variance.
Solution:
◼ Let the original n observations be denoted as x_i
• Then the original mean will be given by: $\bar x ~=~\frac{\sum x_i}{n}$
• Also, the original variance will be given by: ${\sigma^2}_x~=~\frac{\sum(x_i - \bar x)^2}{n}$

Part (i): Adding 'a' to each of the original observation
1. Let the modified n observations be denoted as y_i
• Then the modified observations can be written as:
y₁ = (x₁ + a), y₂ = (x₂ + a), y₃ = (x₃ + a), . . . , y_n = (x_n + a)
2. The modified mean can be calculated as follows:

$\begin{array}{ll}
{}&{\bar y}
& {~=~}& {\frac{\sum y_i}{n}} &{} \\

{}&{}
& {~=~}& {\frac{\sum (x_i + a)}{n}} &{} \\

{}&{}
& {~=~}& {\frac{\left(\sum x_i \right) ~+~ \left( \sum a \right)}{n}} &{} \\

{}&{}
& {~=~}& {\frac{\left(\sum x_i \right) ~+~ na}{n}} &{} \\

{}&{}
& {~=~}& {\frac{\left(\sum x_i \right)}{n}~+~\frac{na}{n}} &{} \\

{}&{}
& {~=~}& {\bar x ~+~ a} &{} \\

\end{array}$

3. Now the modified variance can be calculated as follows:

$\begin{array}{ll}
{}&{{\sigma^2}_y}
& {~=~}& {\frac{\sum(y_i - \bar y)^2}{n}} &{} \\

{}&{}
& {~=~}& {\frac{\sum \left[(x_i + a) - (\bar x + a) \right]^2}{n}} &{} \\

{}&{}
& {~=~}& {\frac{\sum \left[x_i + a - \bar x - a \right]^2}{n}} &{} \\

{}&{}
& {~=~}& {\frac{\sum \left[x_i  - \bar x  \right]^2}{n}} &{} \\

{}&{}
& {~=~}& {{\sigma^2}_x} &{} \\

\end{array}$

◼ That means, there is no change in the variance.

Part (ii): Subtracting 'a' from each of the original observation
1. Let the modified n observations be denoted as y_i
• Then the modified observations can be written as:
y₁ = (x₁ - a), y₂ = (x₂ - a), y₃ = (x₃ - a), . . . , y_n = (x_n - a)
2. The modified mean can be calculated as follows:

$\begin{array}{ll}
{}&{\bar y}
& {~=~}& {\frac{\sum y_i}{n}} &{} \\

{}&{}
& {~=~}& {\frac{\sum (x_i - a)}{n}} &{} \\

{}&{}
& {~=~}& {\frac{\left(\sum x_i \right) ~-~ \left( \sum a \right)}{n}} &{} \\

{}&{}
& {~=~}& {\frac{\left(\sum x_i \right) ~-~ na}{n}} &{} \\

{}&{}
& {~=~}& {\frac{\left(\sum x_i \right)}{n}~-~\frac{na}{n}} &{} \\

{}&{}
& {~=~}& {\bar x ~-~ a} &{} \\

\end{array}$

3. Now the modified variance can be calculated as follows:

$\begin{array}{ll}
{}&{{\sigma^2}_y}
& {~=~}& {\frac{\sum(y_i - \bar y)^2}{n}} &{} \\

{}&{}
& {~=~}& {\frac{\sum \left[(x_i - a) - (\bar x - a) \right]^2}{n}} &{} \\

{}&{}
& {~=~}& {\frac{\sum \left[x_i - a - \bar x + a \right]^2}{n}} &{} \\

{}&{}
& {~=~}& {\frac{\sum \left[x_i  - \bar x  \right]^2}{n}} &{} \\

{}&{}
& {~=~}& {{\sigma^2}_x} &{} \\

\end{array}$

◼ That means, there is no change in the variance.

---

• Based on the results in parts (i) and (ii), we can write:
Adding or subtracting a constant number 'a' from each observation, does not affect the variance.

---

• From step (2) of part (i), we get an useful result. It can be written in 3 steps:
(i) Average of a set of observations is 'A'
(ii) The set is modified by adding a constant 'a' to each observation.
(iii) Then the average of the modified set is (A+a).

---

• From step (2) of part (ii), we get an useful result. It can be written in 3 steps:
(i) Average of a set of observations is 'A'
(ii) The set is modified by subtracting a constant 'a' from each observation.
(iii) Then the average of the modified set is (A-a).

---

Solved example 15.20
The mean and standard deviation of 100 observations were calculated as 40 and 5.1, respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation?
Solution:
Part (i): Finding the correct mean.
1. Let us write the mean for this problem: $\bar x ~=~ \frac{\sum_{i=1}^{i=100}{x_i}}{100}$
2. Ninety nine observations are correct. One observation is wrong. So we will split the above equation. We get:

$\begin{array}{ll}
{}&{{\bar x}_{\text{wrong}}}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{x_i}~+~{x_{\text{wrong}}}}{100}}
&{} \\

{\Rightarrow}&{{\bar x}_{\text{wrong}}}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{x_i}}{100}~+~\frac{x_{\text{wrong}}}{100}}
&{} \\

{\Rightarrow}&{40}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{x_i}}{100}~+~\frac{50}{100}~~ \color {green} {\text{- - - (I)}}}
&{} \\

{\Rightarrow}&{40 - 0.50}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{x_i}}{100}}
&{} \\

{\Rightarrow}&{39.5}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{x_i}}{100}}
&{} \\

{\Rightarrow}&{\sum_{i=1}^{i=99}{x_i}}
& {~=~}& {3950}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
   ♦ We are given that, the wrong mean is 40.
   ♦ We are given that, the wrong observation is 50.

3. Now we can find the correct mean:

$\begin{array}{ll}
{}&{{\bar x}_{\text{correct}}}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{x_i}~+~x_{\text{correct}}}{100}}
&{} \\

{}&{}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{x_i}}{100}~+~\frac{x_{\text{correct}}}{100}}
&{} \\

{}&{}
& {~=~}& {\frac{3950}{100}~+~\frac{40}{100}~~ \color {green} {\text{- - - (I)}}}
&{} \\

{}&{}
& {~=~}& {39.50 + 0.40}
&{} \\

{}&{}
& {~=~}& {39.9}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
   ♦ We use the result $\sum_{i=1}^{i=99}{x_i}~=~3950$ from step (2).
   ♦ We are given that, the correct observation is 40.

Part (ii): Finding the correct variance.
1. We are given that:
Wrong standard deviation is 5.1
• So the wrong variance = (5.1)²
2. Let us write the variance for this problem: $\sigma^2~=~ \frac{\sum_{i=1}^{i=100}{(x_i - \bar x)^2}}{100}$
3. Ninety nine observations are correct. One observation is wrong. So we will split the above equation. We get:

$\begin{array}{ll}
{}&{{\sigma^2}_{\text{wrong}}}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{(x_i - {\bar x}_{\text{wrong}})^2}~+~(x_{\text{wrong}} - {\bar x}_{\text{wrong}})^2}{100}}
&{} \\

{\Rightarrow}&{{\sigma^2}_{\text{wrong}}}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{(x_i - {\bar x}_{\text{wrong}})^2}}{100}~+~\frac{(x_{\text{wrong}} - {\bar x}_{\text{wrong}})^2}{100}}
&{} \\

{\Rightarrow}&{5.1^2}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{(x_i - 40)^2}}{100}~+~\frac{(50 - 40)^2}{100}~~ \color {green} {\text{- - - (I)}}}
&{} \\

{\Rightarrow}&{5.1^2}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{(x_i - 40)^2}}{100}~+~\frac{(10)^2}{100}}
&{} \\

{\Rightarrow}&{5.1^2}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{(x_i - 40)^2}}{100}~+~1}
&{} \\

{\Rightarrow}&{5.1^2~-~1^2}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{\left({x_i}^2 - 80 x_i + 40^2 \right)}}{100}}
&{} \\

{\Rightarrow}&{(5.1+1)(5.1 - 1)}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{\left({x_i}^2 \right)} - 80 \sum_{i=1}^{i=99}{(x_i)} + 99 \times 40^2 }{100}}
&{} \\

{\Rightarrow}&{(6.1)(4.1)}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{\left({x_i}^2 \right)} - 80 (3950) + 99 \times 40^2 }{100}~~ \color {green} {\text{- - - (II)}}}
&{} \\

{\Rightarrow}&{25.01}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{\left({x_i}^2 \right)} - 316000 + 158400 }{100}}
&{} \\

{\Rightarrow}&{25.01}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{\left({x_i}^2 \right)} -157600 }{100}}
&{} \\

{\Rightarrow}&{2501~+~157600}
& {~=~}& {\sum_{i=1}^{i=99}{\left({x_i}^2 \right)}}
&{} \\

{\Rightarrow}&{160101}
& {~=~}& {\sum_{i=1}^{i=99}{\left({x_i}^2 \right)}}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
   ♦ In (1) of part (ii), we wrote the wrong variance.
   ♦ We are given that, the wrong mean is 40.
   ♦ We are given that, the wrong observation is 50.
• Line marked as II:
   ♦ We use the result $\sum_{i=1}^{i=99}{x_i}~=~3950$ from step (2) of part (i).

4. Now we can find the correct variance.

$\begin{array}{ll}
{}&{{\sigma^2}_{\text{correct}}}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{(x_i - {\bar x}_{\text{correct}})^2}~+~(x_{\text{correct}} - {\bar x}_{\text{correct}})^2}{100}}
&{} \\

{}&{}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{(x_i - {\bar x}_{\text{correct}})^2}}{100}~+~\frac{(x_{\text{correct}} - {\bar x}_{\text{correct}})^2}{100}}
&{} \\

{}&{}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{(x_i - 39.9)^2}}{100}~+~\frac{(40 - 39.9)^2}{100}~~ \color {green} {\text{- - - (I)}}}
&{} \\

{}&{}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{\left({x_i}^2 - 2 \times 39.9 \times x_i + 39.9^2 \right)}}{100}~+~\frac{(0.10)^2}{100}}
&{} \\

{}&{}
& {~=~}& {\frac{\sum_{i=1}^{i=99}{\left({x_i}^2 \right)} - 2 \times 39.9 \times \sum_{i=1}^{i=99}{(x_i)} + 99 \times 39.9^2 }{100}~+~\frac{(0.10)^2}{100}}
&{} \\

{}&{}
& {~=~}& {\frac{160101- 2 \times 39.9 \times 3950 + 99 \times 39.9^2 }{100}~+~\frac{(0.10)^2}{100}~~ \color {green} {\text{- - - (II)}}}
&{} \\

{}&{}
& {~=~}& {\frac{160101- 315210
+ 157608.99 }{100}~+~\frac{(0.10)^2}{100}}
&{} \\

{}&{}
& {~=~}& {\frac{2499.99}{100}~+~\frac{(0.10)^2}{100}}
&{} \\

{}&{}
& {~=~}& {\frac{2499.99~+~0.01}{100}}
&{} \\

{}&{}
& {~=~}& {\frac{2500}{100}}
&{} \\

{}&{}
& {~=~}& {25}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
   ♦ In part (i) we calculated the correct mean as 39.9.
   ♦ We are given that, the correct observation is 40.
• Line marked as II:
   ♦ We use the result $\sum_{i=1}^{i=99}{(x_i)^2}~=~160101$ from step (3) of part (ii).
   ♦ We use the result $\sum_{i=1}^{i=99}{x_i}~=~3950$ from step (2) of part (i).

5. Thus we get the correct standard deviation:
${\sigma}_{\text{correct}}~=~ \sqrt{{\sigma^2}_{\text{correct}}} ~=~\sqrt{25}~=~5$

---

Link to a few more miscellaneous examples is given below:

Miscellaneous Exercise

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In the next chapter, we will see probability.

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Chapter 15.6 - Coefficient of Variance

In the previous section, we completed a discussion on the methods for calculating standard deviation. In this section, we will see how standard deviation can be used for the analysis of data.

Let us see an example where we use standard deviation to compare two data sets. It can be written in 7 steps:
1. Two different factories produce cement. They are named as Cement A and Cement B.
• We want to compare the strengths of the two cements.
2. Making cubes:
    ♦ Using Cement A, we make 30 concrete cubes.
    ♦ Using Cement B also, we make 30 concrete cubes.
3. Testing the cubes:
• We place each cube in a compression testing machine and apply a gradually increasing load. We note down the load at which the cube just begins to fail. (some images can be seen here)
• Let the expected load be 25.
4. Differences in the loads:
• For cement A, most of the thirty loads are close to 25. The readings are like: 23, 25, 27, 24 etc.,
• For cement B also, most of the thirty loads are close to 25. But there are some extreme values like, 12, 17, 31 etc.,
5. Comparing the mean values.
    ♦ For cement A, the mean ($\bar{x}_a$) of it’s thirty values is 24.8.
    ♦ For cement B also, the mean ($\bar{x}_b$) of it’s thirty values is 24.8.
6. Based on the mean values, we are tempted to conclude that, both A and B are of the same quality.
7. But in step (4), we saw that B has some extreme values.
• So standard variation (σ_B) of B will be larger than standard variation (σ_A) of A.
• Greater σ means, greater dispersion. So cement B has greater dispersion from the expected value.
• We can conclude that, cement A is of better quality.
(Several different types of tests must be carried out for assessing the quality of materials like cement, steel, sand, electronic capacitors etc., Statistics help us to analyze test results. We will see them in higher classes)

---

• The above example demonstrates how two data sets can be compared.
• Now let us see another case. It can be written in 6 steps:
1. We are given two data sets.
• The first of those two sets contains heights of 45 students.
• The other set contains weights of those 45 students.
• We want to compare the two sets.
2. Calculating standard deviations:
    ♦ Let the standard deviation of the first set be σ₁
    ♦ Let the standard deviation of the second set be σ₂
3. Comparing the two σ values:
    ♦ σ₁ will be in cm.
    ♦ σ₂ will be in kg.
• So we cannot directly compare the two σ values.
4. In such cases, we use coefficient of variation.
• It is calculated using the formula:
$\text{C.V}~=~\frac{\sigma}{\bar{x}} \times 100$
• Where:
    ♦ C.V is the coefficient of variation of the data set.
    ♦ σ is the standard deviation of that data set.
    ♦ $\bar{x}$ is the mean of that data set.
5. We see that:
    ♦ σ comes in the numerator
    ♦ $\bar{x}$ comes in the denominator.
• Since the above σ and $\bar{x}$ are related to the same data set, both will have the same units. So the units will get canceled. We can say that, C.V has no units.
6. In our present example:
• Both (C.V)₁ and (C.V)₂ will not have any units. They are just numbers. So we can compare them.
• A data set having larger C.V is said to be more dispersed than the other.   
• A data set having smaller C.V is said to be more consistent than the other.

---

In some cases, the two data sets will have the same units. But the mean may be different. In such cases also, calculation of C.V becomes necessary.

---

Solved example 15.14
The following table shows the number of workers of two factories and their wages.

Table 15.27

Find the factory in which, wages have greater variability?
Solution:
1. Consider factory A
• There are 5000 workers in this factory. So the data set will contain 5000 wages.
• The mean of those 5000 wages can be easily calculated.
    ♦ Mean is given as: Rs 2500
• Once the mean is calculated, variance can be easily calculated.
    ♦ Variance is given as: 81
    ♦ So standard deviation = √81 = Rs 9        
2. Consider factory B
• There are 6000 workers in this factory. So the data set will contain 6000 wages.
• The mean of those 6000 wages can be easily calculated.
    ♦ Mean is given as: Rs 2500
• Once the mean is calculated, variance can be easily calculated.
    ♦ Variance is given as: 100
    ♦ So standard deviation = √100 = Rs 10
3. Both data sets have the same mean, which is Rs 2500.
4. Based on the above steps, we can write:
• For A:
Dispersion of wages from the mean of 2500 is 9.
• For B:
Dispersion of wages from the mean of 2500 is 10.
5. Thus we get:
Factory B has greater dispersion (variability) of wages than Factory A.

Solved example 15.15
The C.V of two distributions are 60 and 70. Their standard deviations are 21 and 16 respectively. What are their arithmetic means.
Solution:
1. We have the formula:
$\text{C.V}~=~\frac{\sigma}{\bar{x}} \times 100$
• Where:
    ♦ C.V is the coefficient of variation of the data set.
    ♦ σ is the standard deviation of that data set.
    ♦ $\bar{x}$ is the mean of that data set.
2. For the first set, we can write:

$\begin{array}{ll}
{}&{60}
& {~=~}& {\frac{21}{\bar{x}} \times 100}
&{} \\

{\Rightarrow}&{\bar{x}}
& {~=~}& {\frac{21}{60} \times 100}
&{} \\

{}&{}
& {~=~}& {\frac{7}{20} \times 100}
&{} \\

{}&{}
& {~=~}& {7 \times 5}
&{} \\

{}&{}
& {~=~}& {35}
&{} \\

\end{array}$

3. For the second set, we can write:
$\begin{array}{ll}
{}&{70}
& {~=~}& {\frac{16}{\bar{x}} \times 100}
&{} \\

{\Rightarrow}&{\bar{x}}
& {~=~}& {\frac{16}{70} \times 100}
&{} \\

{}&{}
& {~=~}& {\frac{8}{35} \times 100}
&{} \\

{}&{}
& {~=~}& {\frac{8}{7} \times 20}
&{} \\

{}&{}
& {~=~}& {22.85}
&{} \\

\end{array}$

Solved example 15.16
The following values are calculated in respect of heights and weights of the students of a section of class XI:

Table 15.28

Can we say that the weights show greater variation than the heights?
Solution:
1. Consider the data set related to heights.
• Let there be n students. So the data set will contain n heights.
• The mean of those n heights can be easily calculated.
    ♦ Mean is given as: 162.6 cm
• Once the mean is calculated, variance can be easily calculated.
    ♦ Variance is given as: 127.69 cm².
    ♦ So standard deviation = √127.69 = 11.3 cm.
2. Consider the data set related to weights.
• There are n students. So the data set will contain n weights.
• The mean of those n weights can be easily calculated.
    ♦ Mean is given as: 52.36 kg
• Once the mean is calculated, variance can be easily calculated.
    ♦ Variance is given as: 23.1361 kg².
    ♦ So standard deviation = √23.1361 = 4.81 kg.
3. Comparing the results in (1) and (2), we see that, standard deviation is greater for heights.
• But we cannot make such a comparison. This is because, one data set is for heights and the other is for weights.
• To get rid of the units, we must calculate C.V of each data set.
4. We have the formula:
$\text{C.V}~=~\frac{\sigma}{\bar{x}} \times 100$
• Where:
    ♦ C.V is the coefficient of variation of the data set.
    ♦ σ is the standard deviation of that data set.
    ♦ $\bar{x}$ is the mean of that data set.
5. For the heights, we can write:

$\begin{array}{ll}
{}&{\text{C.V}}
& {~=~}& {\frac{11.3}{162.6} \times 100}
&{} \\

{}&{}
& {~=~}& {0.06949 \times 100}
&{} \\

{}&{}
& {~=~}& {6.95}
&{} \\

\end{array}$

6. For the weights, we can write:

$\begin{array}{ll}
{}&{\text{C.V}}
& {~=~}& {\frac{4.81}{52.36} \times 100}
&{} \\

{}&{}
& {~=~}& {0.0918 \times 100}
&{} \\

{}&{}
& {~=~}& {9.18}
&{} \\

\end{array}$

7. We see that:
C.V is greater for weights.
• So we can write:
Weights show greater variability than heights.

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Link to a few more solved examples is given below:

Exercise 15.3

---

In the next section, we will see some miscellaneous examples.

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