16.9 - Miscellaneous Exercise on Axiomatic Probability

In the previous section, we completed a discussion on axiomatic probability. In this section, we will see some miscellaneous examples.

Solved example 16.14
On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability that she visits (i) A before B? (ii) A before B and B before C? (iii) A first and B last? (iv) A either first or second? (v) A just before B?
Solution:
• The first visit can be any one of the four cities.
• The second visit can be any one of the remaining three cities.
• The third visit can be any one of the remaining two cities.
• The fourth visit will be only one remaining city.
• So there are (4 × 3 × 2 × 1) orders in which Veena can visit the four cities. In other words, there are 4! possible orders.
• 4! = 24. So the sample space will contain 24 elements. This is shown below:
S = {
ABCD,    ABDC,    ACBD,    ACDB,    ADBC,    ADCB,
BACD,    BADC,    BCAD,    BCDA,    BDAC,    BDCA,   
CABD,    CADB,    CBAD,    CBDA,    CDAB,    CDBA,
DABC,    DACB,    DBAC,    DBCA,    DCAB,    DCBA
}
• The 24 elements of the sample space can be easily written using the fig.16.3 below:

Fig.16.3

• Since there are 24 elements in S, we can write:
n(S) = 24

Part (i):
1. Let E be the event: She visits A before B.
Let us write the favorable outcomes for E:
• In the first row of S, A comes first in all six cases. So we have 6 from the first row.
• In the second row, B comes first in all six cases. So we have 0 from the second row.
• In the third row, A comes before B in three cases. So we have 3 from the third row.
   ♦ They are: CABD,    CADB,     CDAB
• In the fourth row, A comes before B in three cases. So we have 3 from the third row.
   ♦ They are: DABC,    DACB,     DCAB
• So total number of favorable outcomes =
6 + 0 + 3 + 3 = 12
• We can write: n(E) = 12
2. Since all outcomes in S are equally likely, we get:
$\rm{P(E) = \frac{n(E)}{n(S)} = \frac{12}{24} = \frac{1}{2}}$

Part (ii):
1. Let F be the event: She visits A before B and B before C.
Let us write the favorable outcomes for F:
• In the first row of S, the required order is available in 3 cases. So we have 3 from the first row.
   ♦ They are: ABCD,    ABDC,    ADBC
• In the second row, B comes first in all 6 cases. So we have 0 from the second row.
• In the third row, C comes first in all 6 cases. So we have 0 from the third row.
• In the fourth row, the required order is available in 1 case. So we have 1 from the fourth row.
   ♦ It is: DABC
• So total number of favorable outcomes =
3 + 1 = 4
• We can write: n(F) = 4
2. Since all outcomes in S are equally likely, we get:
$\rm{P(F) = \frac{n(F)}{n(S)} = \frac{4}{24} = \frac{1}{6}}$

Part (iii):
1. Let G be the event: She visits A first and B last.
Let us write the favorable outcomes for G
• In the first row of S, the required order is available in 2 cases. So we have 2 from the first row.
   ♦ They are:  ACDB,    ADCB
• In the second row, B comes first in all 6 cases. So we have 0 from the second row.
• In the third row, C comes first in all 6 cases. So we have 0 from the third row.
• In the fourth row, D comes first in all 6 cases. So we have 0 from the fourth row.
• So total number of favorable outcomes = 2
• We can write: n(G) = 2
2. Since all outcomes in S are equally likely, we get:
$\rm{P(G) = \frac{n(G)}{n(S)} = \frac{2}{24} = \frac{1}{12}}$ 

Part (iv):
1. Let H be the event: She visits A either first or second.
Let us write the favorable outcomes for H
• In the first row of S, the required order is available in all 6 cases. So we have 6 from the first row.
• In the second row, A comes second in 2 cases. So we have 2 from the second row.
   ♦ They are: BACD,    BADC
• In the third row, A comes second in 2 cases. So we have 2 from the third row.
   ♦ They are: CABD,    CADB
• In the fourth row, A comes second in 2 cases. So we have 2 from the fourth row.
   ♦ They are: DABC,    DACB
• So total number of favorable outcomes =
6 + 2 + 2 + 2
• We can write: n(H) = 12
2. Since all outcomes in S are equally likely, we get:
$\rm{P(H) = \frac{n(H)}{n(S)} = \frac{12}{24} = \frac{1}{2}}$

Part (v):
1. Let I be the event: She visits A just before B.
Let us write the favorable outcomes for I
• In the first row of S, the required order is available in 2 cases. So we have 2 from the first row.
   ♦ They are: ABCD,    ABDC
• In the second row, the required order is not available in any of the 6 cases. So we have 0 from the second row.
• In the third row, the required order is available in 2 cases. So we have 2 from the third row.
   ♦ They are: CABD,     CDAB
• In the fourth row, the required order is available in 2 cases. So we have 2 from the fourth row.
   ♦ They are: DABC,     DCAB
• So total number of favorable outcomes =
2 +0 + 2 + 2
• We can write: n(I) = 6
2. Since all outcomes in S are equally likely, we get:
$\rm{P(I) = \frac{n(I)}{n(S)} = \frac{6}{24} = \frac{1}{4}}$

Solved example 16.15
Find the probability that when a hand of 7 cards is drawn from a well shuffled deck of 52 cards, it contains (i) all Kings (ii) 3 Kings (iii) atleast 3 Kings.
Solution:
• Number combinations of 7 cards is $\rm{{}^{52} C_{7}}$.
• So we can write: $\rm{n(S) = {}^{52}C_{7}}$ 

Part (i):
1. We can imagine two boxes.
• The left side box is for the four king cards. The right side box is for the remaining three cards.
• For filling the left side box, we must use only the four king cards. Number of possible combinations of four king cards, taking all four at a time is $\rm{{}^{4} C_{4}}$. So this box  can be filled in $\rm{{}^{4} C_{4}}$ ways.
• For filling the right side box, we must not use any of the four king cards. So the right side box can be filled in $\rm{{}^{48} C_{3}}$ ways.
• The two boxes can be filled together in
$\rm{{}^{4} C_{4}~\times~{}^{48} C_{3}}$ ways.
• Thus the number of combinations with all kings is:
$\rm{{}^{4} C_{4}~\times~{}^{48} C_{3}}$
• We can write:
If A is the event: “getting all the kings”, then:
$\rm{n(A) = {}^{4} C_{4}~\times~{}^{48} C_{3}}$
2. Since all outcomes in S are equally likely, we get:
$\rm{P(A) = \frac{n(A)}{n(S)} = \frac{{}^{4} C_{4}~\times~{}^{48} C_{3}}{{}^{52} C_{7}} = \frac{1}{7735}}$

Part (ii):
1. We can imagine two boxes.
• The left side box is for the three king cards. The right side box is for the remaining four cards.
• For filling the left side box, we must use only the four king cards. Number of possible combinations of four king cards, taking three at a time is $\rm{{}^{4} C_{3}}$. So this box  can be filled in $\rm{{}^{4} C_{3}}$ ways.
• For filling the right side box, we must not use any of the four king cards. Because, the combination must contain exactly three kings. So the right side box can be filled in $\rm{{}^{48} C_{4}}$ ways.
• The two boxes can be filled together in
$\rm{{}^{4} C_{3}~\times~{}^{48} C_{4}}$ ways.
• Thus the number of combinations with all kings is:
$\rm{{}^{4} C_{3}~\times~{}^{48} C_{4}}$
• We can write:
If B is the event: “getting exactly three kings”, then:
$\rm{n(B) = {}^{4} C_{3}~\times~{}^{48} C_{4}}$
2. Since all outcomes in S are equally likely, we get:
$\rm{P(B) = \frac{n(B)}{n(S)} = \frac{{}^{4} C_{3}~\times~{}^{48} C_{4}}{{}^{52} C_{7}} = \frac{9}{1547}}$

Part (iii):
1. Consider the two events A and B that we saw in parts (i) and (ii) respectively above.
   ♦ In A, each combination has exactly 4 kings.
   ♦ In B, each combination has exactly 3 kings.
• So A and B are mutually exclusive events.
2. Consider the set (A∪B).
When the outcome is from (A∪B), two things are possible:
(i) The outcome is from A. Then the condition “atleast 3 kings” is satisfied.
(ii) The outcome is from B. Then also the condition “atleast 3 kings” is satisfied.
3. So we want P(A∪B)
• We have: P(A∪B) = P(A) + P(B)
• Substituting the known values, we get:
P(A∪B) = $\frac{1}{7735} + \frac{9}{1547} = \frac{46}{7735}$

Solved example 16.16
If A, B, C are three events associated with a random experiment, prove that
P(A∪B∪C) = P(A) + P(B) +P(C) − P(A∩B) − P(A∩C) – P(B∩C) + P(A∩B∩C)
Solution:
1. Consider the LHS of the given expression.
• We can think of a new event (B∪C)
• Let E = (B∪C)
2. Now the LHS becomes:

$\begin{array}{ll}
{}&{\rm{P(A∪B∪C)}}
& {~=~}& {\rm{P(A∪E)}} &{} \\

{}&{}
& {~=~}& {\rm{P(A) + P(E) - P(A∩E)}} &{} \\

\end{array}$

3. Next step is to simplify the second term in the RHS of (2). The second term is P(E). We can write:

$\begin{array}{ll}
{}&{\rm{P(E)}}
& {~=~}& {\rm{P(B∪C)}} &{} \\

{}&{}
& {~=~}& {\rm{P(B) + P(C) - P(B∩C)}} &{} \\

\end{array}$

4. Next step is to simplify the third term in the RHS of (2). The third term is P(A∩E). We can write:

$\begin{array}{ll}
{}&{\rm{A∩E}}
& {~=~}& {\rm{A∩(B∪C)}}
&{} \\

{\Rightarrow}&{\rm{A∩E}}
& {~=~}& {\rm{(A∩B)∪(A∩C)~\color{green}{\text{- - - I}}}}
&{} \\

{\Rightarrow}&{\rm{P(A∩E)}}
& {~=~}& {\rm{P \Bigl((A∩B)∪(A∩C) \Bigr)}}
&{} \\

{\Rightarrow}&{\rm{P(A∩E)}}
& {~=~}& {\rm{P(A∩B)~+~P(A∩C)~-~P \Bigl((A∩B)∩(A∩C) \Bigr)}}
&{} \\

{\Rightarrow}&{\rm{P(A∩E)}}
& {~=~}& {\rm{P(A∩B)~+~P(A∩C)~-~P(A∩B∩C)~\color{green}{\text{- - - II}}}}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
In this line we use the “distribution property of intersection of sets over the union”
• Line marked as II:
Here we use the fact that (A∩B)∩(A∩C) = (A∩B∩C)
5. Now we can make the substitutions:
   ♦ From (3), we have the substitute for P(E)  
   ♦ From (4), we have the substitute for P(A∩E)  
• Making these substitutions in (2), we get:
P(A∪B∪C)
= P(A) + P(B) + P(C) - P(B∩C) - P(A∩B) - P(A∩C) + P(A∩B∩C)

Solved example 16.17
In a relay race there are five teams A, B, C, D and E.
(a) What is the probability that A, B and C finish first, second and third, respectively.
(b) What is the probability that A, B and C are first three to finish (in any order)
(Assume that all finishing orders are equally likely)
Solution:
• Five teams can finish in 5! ways.
• So the number of elements in S = 5! = 120
• We can write: n(S) = 120

Part (i):
1. Let G be the event: A, B and C finish first, second and third respectively.
• There are only two possible outcomes for such a finish. They are:
ABC, D, E and ABC, E, D
• So we can write: n(G) = 2
2. Since all outcomes in S are equally likely, we get:
$\rm{P(G) = \frac{n(G)}{n(S)} = \frac{2}{120} = \frac{1}{60}}$
Part (ii):
1. Let H be the event: A, B and C are the first three to finish in any order.
• A, B and C can arrange among themselves in 3! ways.
   ♦ We have, 3! = 3 × 2 = 6
• For each of those 6 ways, D and E can arrange among themselves in 2! ways.
   ♦ We have, 2! = 2 × 1 = 2 ways.
• So the number of favorable outcomes = 6 × 2 = 12
• We can write: n(H) = 12
2. Since all outcomes in S are equally likely, we get:
$\rm{P(H) = \frac{n(H)}{n(S)} = \frac{12}{120} = \frac{1}{10}}$

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Link to a few more solved examples is given below:

Miscellaneous Exercise

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In the next section, we will see Appendix A.

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16.8 Solved Examples on Axiomatic Probability

In the previous section, we saw how to calculate the Probability of the event "not A". We saw two solved examples also. In this section, we will see two more solved examples.

Solved example 16.12
Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that
(a) Both Anil and Ashima will not qualify the examination.
(b) Atleast one of them will not qualify the examination and
(c) Only one of them will qualify the examination.
Solution:
First we will write two basic points. We will number them as 𝛼 and β:
Point 𝛼:
This can be written in 6 steps:
1. Consider the Venn diagram shown in fig.16.2(a) below. It is already familiar to us.

Fig.6.2

2. (Red ∪ Blue) gives set A.
In our present case, A is the event: Anil qualifies in the examination.
3. (Green ∪ Blue) gives set B.
In our present case, B is the event: Ashima qualifies in the examination.
4. Blue is the set (A∩B).
In our present case, (A∩B) is the event: Both Anil and Ashima qualify in the examination.
5. (Red ∪ Blue ∪ Green) gives the set (A∪B).
In our present case, (A∪B) is the event: Anil or Ashima qualifies in the examination.
6. Consider the portion outside (A∪B), but inside the rectangle. That portion gives the set (A∪B)’. It is the portion shown in yellow color in fig.16.2(b) above.
In our present case, (A∪B)’ is the event: Both Anil and Ashima does not qualify in the examination.

Point β:
This can be written in 6 steps:
1. Consider the rectangle S. Imagine that, there are a large number of elements dispersed inside the rectangle. Each of those elements is an outcome.
2. Consider all the outcomes in S. Some of those outcomes are present within (Red ∪ Blue).
• If the outcome of the experiment is from this region, we say that:
Anil has qualified.
• The probability for the outcome to be from this region is given in the question. P(A) = 0.05
3. Consider all the outcomes in S. Some of those outcomes are present within (Green ∪ Blue).
• If the outcome of the experiment is from this region, we say that:
Ashima has qualified.
• The probability for the outcome to be from this region is given in the question. P(B) = 0.1
4. Consider all the outcomes in S. Some of those outcomes are present within Blue.
• If the outcome of the experiment is from this region, we say that:
Both Anil and Ashima have qualified.
• The probability for the outcome to be from this region is given in the question. P(A∩B) = 0.02
5. A and B are not disjoint sets. This can be proved in 3 steps:
(i) If A and B are disjoint sets, (A∩B) = Φ
(ii) Probability of Φ is zero.
(iii) But according to the question, P(A∩B) is not zero. It is 0.02
6. We said that, a large number of outcomes are dispersed within the rectangle S.
• In our present case,
    ♦ we do not know how many such outcomes are there.
    ♦ we do not know the probabilities of each of those outcomes.
• If we knew the number of outcomes and their probabilities, we could calculate P(A), P(B) etc.,
• But as the reader may have already noted, in our present case, we do not need them because, P(A), P(B) and P(A∩B) are already given.  

---

Now we can answer the questions.
Part (i): Both Anil and Ashima will not qualify.
1. Consider the region (A∪B). It is made up of three regions: red, blue and green.
• If the outcome is from the red region, Anil qualifies. So “Both Anil and Ashima will not qualify” is not satisfied.
• If the outcome is from the blue region, Anil and Ashima qualifies. So “Both Anil and Ashima will not qualify” is not satisfied.
• If the outcome is from the green region, Ashima qualifies. So “Both Anil and Ashima will not qualify” is not satisfied.
• It is clear that, we must discard (A∪B).
2. Consider the region outside (A∪B), but inside the rectangle.
• We know that, such a region is the compliment of set (A∪B). We denote it as (A∪B)’. It is the yellow region of the Venn diagram in fig.16.2(b) above.
• If the outcome is from (A∪B)', we say that:
Both Anil and Ashima will not qualify.
3. The probability for the outcome to be from (A∪B)' is: P(A∪B)'
• So our aim is to find P(A∪B)'
4. It is clear that, (A∪B) and (A∪B)’ are mutually exclusive and exhaustive events.
• We can write:
(A∪B)∪(A∪B)' = S.
• Based on this, we can write the calculations as follows:
$\begin{array}{ll}
{}&{\rm{(A \cup B) \cup (A \cup B)'}}
& {~=~}& {\rm{S}}
&{} \\

{\Rightarrow}&{\rm{P \Bigl((A \cup B) \cup (A \cup B)' \Bigr)}}
& {~=~}& {\rm{P(S)}}
&{} \\

{\Rightarrow}&{\rm{P(A \cup B) + P(A \cup B)'}}
& {~=~}& {\rm{P(S)~\color{green}{\text{- - - I}}}}
&{} \\

{\Rightarrow}&{\rm{P(A) + P(B) - P(A \cap B) + P(A \cup B)'}}
& {~=~}& {\rm{P(S)~\color{green}{\text{- - - II}}}}
&{} \\

{\Rightarrow}&{\rm{0.05 + 0.1 - 0.02 + P(A \cup B)'}}
& {~=~}& {\rm{1}}
&{} \\

{\Rightarrow}&{\rm{0.13 + P(A \cup B)'}}
& {~=~}& {\rm{1}}
&{} \\

{\Rightarrow}&{\rm{P(A \cup B)'}}
& {~=~}& {\rm{1 - 0.13}}
&{} \\

{\Rightarrow}&{\rm{P(A \cup B)'}}
& {~=~}& {\rm{0.87}}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
In this line we use the formula:
P(E∪F) = P(E) + P(F)
Where E and F are disjoint sets.
• Line marked as II:
In this line we use the formula:
P(E∪F) = P(E) + P(F) - P(E∩F)
Where E and F are not disjoint sets.

Part (ii): Atleast one of them will not qualify the examination
1. Consider the blue region.
If the outcome is from this region, it means that both Anil and Ashima qualifies for the examination. So we have to discard this region.
2. Consider the region outside blue but inside the rectangle. This region is (A∩B)’
• This (A∩B)’ is made up of three regions:
(i) Red region (ii) Green region (iii) yellow region.
3. Let us examine each of the three regions.
(i) If the outcome is from the red region, then Anil qualifies but Ashima does not qualify. So “atleast one of them will not qualify” is satisfied.
(ii) If the outcome is from the green region, then Ashima qualifies but Anil does not qualify. So “atleast one of them will not qualify” is satisfied.
(iii) If the outcome is from yellow region, then both do not qualify. So “atleast one of them will not qualify” is satisfied.
4. So (A∩B)’ is our required region.
• The probability for the outcome to be from this region is: P(A∩B)’
5. We have:

$\begin{array}{ll}
{}&{\rm{P(A \cap B)'}}
& {~=~}& {\rm{1 - P(A \cap B)}} &{} \\

{}&{}
& {~=~}& {\rm{1 - 0.02}} &{} \\

{}&{}
& {~=~}& {\rm{0.98}} &{} \\

\end{array}$

Part (iii): Only one of them will qualify the examination.
1. Let us examine each region in fig.16.2 above.
(i) The red region.
If the outcome is from red, then only Anil qualifies. So this region can be considered for our answer.
(ii) The blue region.
If the outcome is from blue, then both qualify. So this region cannot be considered for our answer.
(iii) Green region.
If the outcome is from green, then only Ashima qualifies. So this region can be considered for our answer.
(iv) Yellow region.
If the outcome is from yellow , then neither Anil nor Ashima qualifies. So this region cannot be considered for our answer.
2. Based on the above step, we can write:
The only regions than can be considered are: red and green.
3. We can create a new set: (red ∪ green)
If the outcome is from this union, there are two possibilities:
(i) outcome is from red.
Then only Anil qualifies. So “only one of them will qualify” is satisfied.
(ii) outcome is from green.
Then only Ashima qualifies. So “only one of them will qualify” is satisfied.
4. So (red ∪ green) is our required region.
    ♦ Red is (A-B)
    ♦ Green is (B-A)
• So (A-B)∪(B-A) is our required region.
5. Probability for the outcome to be from this region is: $P \Bigl((A-B) \cup (B-A) \Bigr)$
• (A-B) and (B-A) are disjoint sets. So we can write:
$P \Bigl((A-B) \cup (B-A) \Bigr) = P \Bigl((A-B)\Bigr) + P \Bigl((B-A)\Bigr)$
• So we have to calculate $P \Bigl((A-B)\Bigr) ~\text{and}~ P \Bigl((B-A)\Bigr)$
6. First we will calculate $P \Bigl((A-B)\Bigr)$

$\begin{array}{ll}
{}&{\rm{A}}
& {~=~}& {\rm{(A-B) \cup (A \cap B)~\color{green}{\text{- - - I}}}}
&{} \\

{\Rightarrow}&{\rm{P(A)}}
& {~=~}& {\rm{P \Bigl((A-B) \cup (A \cap B)\Bigr)}}
&{} \\

{\Rightarrow}&{\rm{P(A)}}
& {~=~}& {\rm{P \Bigl((A-B)\Bigr) ~+~ P \Bigl((A \cap B)\Bigr)~\color{green}{\text{- - - II}}}}
&{} \\

{\Rightarrow}&{\rm{0.05}}
& {~=~}& {\rm{P \Bigl((A-B)\Bigr) ~+~ 0.02}}
&{} \\

{\Rightarrow}&{\rm{P \Bigl((A-B)\Bigr)}}
& {~=~}& {\rm{0.05~-~ 0.02}}
&{} \\

{\Rightarrow}&{\rm{P \Bigl((A-B)\Bigr)}}
& {~=~}& {\rm{0.03}}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
Set A is the union of red and blue.
• Line marked as II:
We are able to simply add the individual probabilities because, (A-B) and (A∩B) are disjoint sets.

7. Next we will calculate $P \Bigl((B-A)\Bigr)$

$\begin{array}{ll}
{}&{\rm{B}}
& {~=~}& {\rm{(B-A) \cup (A \cap B)~\color{green}{\text{- - - I}}}}
&{} \\

{\Rightarrow}&{\rm{P(B)}}
& {~=~}& {\rm{P \Bigl((B-A) \cup (A \cap B)\Bigr)}}
&{} \\

{\Rightarrow}&{\rm{P(B)}}
& {~=~}& {\rm{P \Bigl((B-A)\Bigr) ~+~ P \Bigl((A \cap B)\Bigr)~\color{green}{\text{- - - II}}}}
&{} \\

{\Rightarrow}&{\rm{0.1}}
& {~=~}& {\rm{P \Bigl((B-A)\Bigr) ~+~ 0.02}}
&{} \\

{\Rightarrow}&{\rm{P \Bigl((B-A)\Bigr)}}
& {~=~}& {\rm{0.1~-~ 0.02}}
&{} \\

{\Rightarrow}&{\rm{P \Bigl((B-A)\Bigr)}}
& {~=~}& {\rm{0.08}}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
Set B is the union of green and blue.
• Line marked as II:
We are able to simply add the individual probabilities because, (B-A) and (A∩B) are disjoint sets. 

8. Substituting the results from (6) and (7) in (5), we get:
$P \Bigl((A-B) \cup (B-A) \Bigr) = 0.03 + 0.08 = 0.11$

Solved example 16.13
A committee of two persons is selected from two men and two women. What is the probability that the committee will have (a) no man? (b) one man? (c) two men?
Solution:
◼ There are two men (M1 & M2) and two women (W1 & W2)
    ♦ So there is a total of four persons
◼ From that four, two persons can be selected in $\rm{{}^4 C_2}$ ways.
    ♦ So the number of possible outcomes = $\rm{{}^4 C_2}$
    ♦ We can write: n(S) = $\rm{{}^4 C_2}$
    ♦ All the $\rm{{}^4 C_2}$ outcomes are equally likely.
    ♦ Some of those outcomes are: (M1, W1), (W2, M1), etc.,
• Now we can do the calculations:
Part (i):
1. Let A be the event: Getting an outcome with no man.  
2. Since there is to be no man, we must not consider M1 and M2 while making the selections.
• That means, we must consider W1 and W2 only.
• Two women can be selected from two women in $\rm{{}^2 C_2}$ ways.
• So n(A) = $\rm{{}^2 C_2}$
3. Since all outcomes are equally likely, we get:
$\rm{P(A) = \frac{n(A)}{n(S)} = \frac{{}^2 C_2}{{}^4 C_2} = \frac{1}{6}}$

Part (ii):
1. Let B be the event: Getting an outcome with one man.
2. Since there is to be exactly one man, the other person in the committee will be woman.
• One man can be selected from two men in $\rm{{}^2 C_1}$ ways.
• One woman can be selected from two women in $\rm{{}^2 C_1}$ ways.
• Together, they can be selected in $\rm{{}^2 C_1 \times {}^2 C_1}$ ways.
• So n(B) = $\rm{{}^2 C_1 \times {}^2 C_1}$
3. Since all outcomes are equally likely, we get:
$\rm{P(B) = \frac{n(B)}{n(S)} = \frac{{}^2 C_1 \times {}^2 C_1}{{}^4 C_2} = \frac{2 \times 2}{6} = \frac{2}{3}}$

Part (iii):
1. Let C be the event: Getting an outcome with two men.  
2. Since there is to be two men, we must not consider W1 and W2 while making the selections.
• That means, we must consider M1 and M2 only.
• Two men can be selected from two men in $\rm{{}^2 C_2}$ ways.
• So n(C) = $\rm{{}^2 C_2}$
3. Since all outcomes are equally likely, we get:
$\rm{P(C) = \frac{n(C)}{n(S)} = \frac{{}^2 C_2}{{}^4 C_2} = \frac{1}{6}}$

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Link to a few more solved examples is given below:

Exercise 16.3

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

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16.4 - Axiomatic Approach to Probability

In the previous section, we saw Exhaustive events. In this section, we will see Axiomatic approach to probability.

• In our earlier classes, we have seen how probability can be calculated in simple cases.
• We calculated probability as a fraction. It was a proper fraction. That is., numerator will be less than the denominator. If a proper fraction is converted to decimal form, the value on the left side of the decimal point will be zero.
• In the denominator, we gave total number of possible outcomes. In the numerator, we gave the total number of favorable outcomes.
• In our present discussion, we will see three axioms developed by the Russian mathematician A.N. Kolmogorov. Those axioms will enable us to calculate probability in complex problems.
• Axiom means a statement or rule, which is self-evident. Such a statement/rule can be a useful basis for more advanced mathematical calculations.

Axiom 1: For any event A, P(A) ≥ 0.
• This can be explained in 5 steps:
(i) The probability any event A to occur will be greater than or equal to zero.
    ♦ If P(A) is zero, A will never occur.
    ♦ If P(A) is large, there is a higher probability for A to occur.
    ♦ If P(A) is small, there is a lower probability for A to occur.
(ii) We saw that, probability is always a proper fraction. So the largest possible value for P(A) is 1. When the probability is 1, the event will surely occur.
(iii) Also, the smallest possible value for P(A) is zero. When the probability is zero, the event will never occur.
(iv) Since P(A) is a proper fraction, “zero value” is obtained when the numerator is zero. This happens when there are no favorable outcomes.
(v) Since P(A) is a proper fraction, “1 value” is obtained when the numerator is same as the denominator. This happens when number of favorable outcomes is same as the total number of outcomes.

Axiom 2:
For any experiment, P(S) = 1
• This can be explained in 2 steps:
(i) We know that, any event A can be denoted as a set A. The elements of set A will be the favorable outcomes of event A.
(ii) The sample space of an experiment is also a set. We denote it as S.
• So this S can be considered as an event. Whenever we perform that experiment, we will surely obtain an element of S. Based on this, we can say: Probability for S to occur is 1.

Axiom 3:
If A and B are two mutually exclusive events then P(A∪B) = P(A) + P(B)
• This can be explained using an example. It can be written in 5 steps:
(i) First we will assign the sets.
• Let set S of an experiment have 6 elements.
• Let event A of that experiment have 3 elements.
• Let event B of that experiment have 2 elements.
(ii) Now we will write the probabilities of the individual events:
• We get:
    ♦ P(A) = 3/6 = 1/2
    ♦ P(B) = 2/6 = 1/3
(iii) Given that A and B are disjoint sets. So A∪B will have 5 elements.
• So P(A∪B) = 5/6
(iv) Let us add the individual probabilities:
P(A) + P(B) = 1/2 + 1/3 = 5/6
• This is same as the result in (iii)
(v) Comparing the results in (iii) and (iv), we can write:
P(A∪B) = P(A) + P(B)

---

Using the above axioms, we can derive four important results.
Result 1:
This can be written in 3 steps:
1. Suppose that, B is a null set. That means, no outcome in S can be used to define the event B.
(For example, when a die is thrown once, the event of “getting number 7” is a null set)
• So we can write: B = Ф
2. Applying axiom 3, we get:
$\begin{array}{ll}
{}&{P(A \cup B)}
& {~=~}& {P(A) + P(B)}
&{} \\

{\Rightarrow}&{P(A \cup \phi)}
& {~=~}& {P(A) + P(\phi)~~ \color {green} {\text{- - - (I)}}}
&{} \\

{\Rightarrow}&{P(A)}
& {~=~}& {P(A) + P(\phi)~~ \color {green} {\text{- - - (II)}}}
&{} \\

{\Rightarrow}&{P(\phi)}
& {~=~}& {0}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
In this line, we are able to apply axiom 3 because, any set A and the null set are disjoint sets
• Line marked as II:
In this line, we use the rule:
Union of any set A with the null set will give A
3. Based on the above calculations, we can write:
If there are no outcomes for an event, then the probability for that event to occur is zero.

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Result 2:
This can be written in 4 steps:
1. Let S be the sample space of an experiment.
• Also, let 𝜔₁, 𝜔₂, 𝜔₃, . . . , 𝜔_n be the n elements of S.
• Then we can write: S= {𝜔₁, 𝜔₂, 𝜔₃, . . . , 𝜔_n}
2. So S is the union of n singleton sets:
S = {𝜔₁} ∪ {𝜔₂} ∪ {𝜔₃} ∪ . . . ∪ {𝜔_n}
3. Applying axiom 3, we get:

$\begin{array}{ll}
{}&{\rm{P \left(\lbrace \omega_1 \rbrace \cup \lbrace \omega_2 \rbrace \cup \lbrace \omega_3 \rbrace \cup ~.~.~.~ \cup \lbrace \omega_n \rbrace \right)}}
& {~=~}& {\rm{P \left(\lbrace \omega_1 \rbrace \right) + P \left(\lbrace \omega_2 \rbrace \right) + P \left(\lbrace \omega_3 \rbrace \right) ~+~.~.~.~+~P \left(\lbrace \omega_n \rbrace \right)}}
&{} \\

{\Rightarrow}&{\rm{P(S)}}
& {~=~}& {\rm{P \left(\lbrace \omega_1 \rbrace \right) + P \left(\lbrace \omega_2 \rbrace \right) + P \left(\lbrace \omega_3 \rbrace \right) ~+~.~.~.~+~P \left(\lbrace \omega_n \rbrace \right)}}
&{} \\

{\Rightarrow}&{\rm{1}}
& {~=~}& {\rm{P \left(\lbrace \omega_1 \rbrace \right) + P \left(\lbrace \omega_2 \rbrace \right) + P \left(\lbrace \omega_3 \rbrace \right) ~+~.~.~.~+~P \left(\lbrace \omega_n \rbrace \right)}}
&{} \\

\end{array}$

◼ Remarks:
From axiom 2, we know that: P(S) = 1

4. So we can write:
Sum of the individual probabilities of all individual outcomes of an experiment will be 1.

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Result 3:
This can be written in 5 steps:
1. From result 2, we know that:
P({𝜔₁}) + P({𝜔₂}) + P({𝜔₃}) + . . . + P({𝜔_n}) = 1
• Note that, the sum on the RHS is 1.
2. Each term on the LHS is a proper fraction.
• Since the sum is 1, each term must be either less than 1 or equal to 1.
3. But axiom 1 tells us that:
Each term will be either greater than zero or equal to zero.
4. Combining (2) and (3), we get:
0 ≤ Any term ≤ 1
• Mathematically, we write this as:
0 ≤ P({𝜔_i}) ≤ 1
5. So we can write:
Take any outcome from S. That outcome can be considered as a simple event (event with only one outcome). The probability of that event will be greater than or equal to zero and at the same time, less than or equal to 1.

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Result 4:
This can be written in 4 steps:
1. Let A be an event in an experiment.
• Also, let 𝜔₁, 𝜔₂, 𝜔₃, . . . , 𝜔_n be the n elements of A.
• Then we can write: A = {𝜔₁, 𝜔₂, 𝜔₃, . . . , 𝜔_n}
2. So A is the union of n singleton sets:
A = {𝜔₁} ∪ {𝜔₂} ∪ {𝜔₃} ∪ . . . ∪ {𝜔_n}
3. Applying axiom 3, we get:

$\begin{array}{ll}
{}&{\rm{P \left(\lbrace \omega_1 \rbrace \cup \lbrace \omega_2 \rbrace \cup \lbrace \omega_3 \rbrace \cup ~.~.~.~ \cup \lbrace \omega_n \rbrace \right)}}
& {~=~}& {\rm{P \left(\lbrace \omega_1 \rbrace \right) + P \left(\lbrace \omega_2 \rbrace \right) + P \left(\lbrace \omega_3 \rbrace \right) ~+~.~.~.~+~P \left(\lbrace \omega_n \rbrace \right)}}
&{} \\

{\Rightarrow}&{\rm{P(A)}}
& {~=~}& {\rm{P \left(\lbrace \omega_1 \rbrace \right) + P \left(\lbrace \omega_2 \rbrace \right) + P \left(\lbrace \omega_3 \rbrace \right) ~+~.~.~.~+~P \left(\lbrace \omega_n \rbrace \right)}}
&{} \\

{\Rightarrow}&{\rm{P(A)}}
& {~=~}& {\rm{\sum{P \left(\lbrace \omega_i \rbrace \right)}}}
&{} \\

\end{array}$

4. So we can write:
Sum of the individual probabilities of all individual outcomes of an event will be the probability of that event.

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◼ Based on the three axioms and four results, we can write:
Any experiment must satisfy two conditions:
Condition 1:
Take any individual outcome of an experiment. The probability of that outcome will be greater than or equal to zero, but at the same time less than or equal to 1.

• Mathematically, we can write this as:
$\rm{0 \le P(\lbrace \omega_i\rbrace) \le 1,~Where~ \omega_i \in S}$
Condition 2:
Take the sum of the probabilities of all individual outcomes in S. That sum will be 1.

• Mathematically, we can write this as:
$\rm{\sum{P(\lbrace \omega_i\rbrace)} = 1,~Where~ \omega_i \in S}$

• Let us see an example. It can be written in steps:
1. Consider the experiment of tossing a coin once.
We know that, S = {H,T}
2. Let us assign individual probabilities for the two outcomes:
    ♦ P({H}) = ½
    ♦ P({T}) = ½
3. Does the above probability values satisfy the two conditions?
(i) We have: 0 ≤ ½ ≤ 1
So condition 1 is satisfied.
(ii) We have: (½ + ½) = 1
So condition 2 is also satisfied.
4. We know that exact ½ cannot be obtained even if we repeat the experiment 50000 times. What if we repeat the experiment only 20 times and get the following values?
    ♦ P({H}) = 1/4
    ♦ P({T}) = 3/4
5. Does the above probability values satisfy the two conditions?
(i) We have:
    ♦ 0 ≤ 1/4 ≤ 1
    ♦ 0 ≤ 3/4 ≤ 1
So condition 1 is satisfied.
(ii) We have: (1/4 + 3/4) = 1
So condition 2 is also satisfied.
6. We see that ¼ and ¾ also satisfy the two conditions.
• In fact, there are infinite number of possible values such that:
    ♦ P({H}) = p
    ♦ P({T}) = (1- p)
Where 0 ≤ p ≤ 1

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Now we will see a solved example:

Solved example 16.9
Let a sample space be S = {𝜔₁, 𝜔₂, 𝜔₃, . . . , 𝜔₆}. Which of the following assignments of probabilities to each outcome are valid ?

$\begin{array}{cc}
{\textbf{Outcomes}}&{\mathbf{\omega_1}}
& {\mathbf{\omega_2}}& {\mathbf{\omega_3}}
&{\mathbf{\omega_4}} &{\mathbf{\omega_5}} &{\mathbf{\omega_6}}\\

{\textbf{(a)}}&{\frac{1}{6}}
& {\frac{1}{6}}& {\frac{1}{6}}
&{\frac{1}{6}} &{\frac{1}{6}} &{\frac{1}{6}}\\

{\textbf{(b)}}&{1}
& {0}& {0}
&{0} &{0} &{0}\\

{\textbf{(c)}}&{\frac{1}{8}}
& {\frac{2}{3}}& {\frac{1}{3}}
&{\frac{1}{3}} &{- \frac{1}{4}} &{- \frac{1}{3}}\\

{\textbf{(d)}}&{\frac{1}{12}}
& {\frac{1}{12}}& {\frac{1}{6}}
&{\frac{1}{6}} &{\frac{1}{6}} &{\frac{3}{2}}\\

{\textbf{(e)}}&{0.1}
& {0.2}& {0.3}
&{0.4} &{0.5} &{0.6}\\

\end{array}$

Solution:
We have two conditions to check whether the probabilities assigned to the outcomes are valid.
Condition 1:
$\rm{0 \le P(\lbrace \omega_i\rbrace) \le 1,~Where~ \omega_i \in S}$
Condition 2:
$\rm{\sum{P(\lbrace \omega_i\rbrace)} = 1,~Where~ \omega_i \in S}$

Let us check each case:
Part (a):
• Condition 1 is satisfied because, all probability values are greater than or equal to 0 and at the same time, less than or equal to 1.
• Condition 2 is satisfied because:
$\frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6}~=~1$
• Both conditions are satisfied. So the assigned probabilities are valid.

Part (b):
• Condition 1 is satisfied because, all probability values are greater than or equal to 0 and at the same time, less than or equal to 1.
• Condition 2 is satisfied because:
1 + 0 + 0 + 0 + 0 + 0 = 1
• Both conditions are satisfied. So the assigned probabilities are valid.

Part (c):
• Condition 1 is not satisfied because, P({𝜔₅}) and P({𝜔₆}) are less than 0.
• Since condition 1 is not satisfied, there is no need to check condition 2.
• Condition 1 is not satisfied. So the assigned probabilities are not valid.

Part (d):
• Condition 1 is not satisfied because, P({𝜔₆}) is greater than 1.
• Since condition 1 is not satisfied, there is no need to check condition 2.
• Condition 1 is not satisfied. So the assigned probabilities are not valid.

Part (e):
• Condition 1 is satisfied because, all probability values are greater than or equal to 0 and at the same time, less than or equal to 1.
• Condition 2 is not satisfied because:
0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 = 2.1 ≠ 1
• Condition 2 is not satisfied. So the assigned probabilities are not valid.

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In the next section, we will see Probability of an event.

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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.

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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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