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:
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The first visit can be any one of the four cities.
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The second visit can be any one of the remaining three cities.
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The third visit can be any one of the remaining two cities.
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The fourth visit will be only one remaining city.
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So there are (4 × 3 × 2 × 1) orders in which Veena can visit the four cities. In other words, there are 4! possible orders.
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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
}
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The 24 elements of the sample space can be easily written using the fig.16.3 below:
 |
| Fig.16.3 |
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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:
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In the first row of S, A comes first in all six cases. So we have 6 from the first row.
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In the second row, B comes first in all six cases. So we have 0 from the second row.
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In the third row, A comes before B in three cases. So we have 3 from the third row.
♦ They are: CABD, CADB, CDAB
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In the fourth row, A comes before B in three cases. So we have 3 from the third row.
♦ They are: DABC, DACB, DCAB
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So total number of favorable outcomes =
6 + 0 + 3 + 3 = 12
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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:
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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
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In the second row, B comes first in all 6 cases. So we have 0 from the second row.
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In the third row, C comes first in all 6 cases. So we have 0 from the third row.
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In the fourth row, the required order is available in 1 case. So we have 1 from the fourth row.
♦ It is: DABC
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So total number of favorable outcomes =
3 + 1 = 4
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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
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In the second row, B comes first in all 6 cases. So we have 0 from the second row.
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In the third row, C comes first in all 6 cases. So we have 0 from the third row.
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In the fourth row, D comes first in all 6 cases. So we have 0 from the fourth row.
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So total number of favorable outcomes = 2
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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
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In the first row of S, the required order is available in all 6 cases. So we have 6 from the first row.
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In the second row, A comes second in 2 cases. So we have 2 from the second row.
♦ They are: BACD, BADC
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In the third row, A comes second in 2 cases. So we have 2 from the third row.
♦ They are: CABD, CADB
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In the fourth row, A comes second in 2 cases. So we have 2 from the fourth row.
♦ They are: DABC, DACB
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So total number of favorable outcomes =
6 + 2 + 2 + 2
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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
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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
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In the second row, the required order is not available in any of the 6 cases. So we have 0 from the second row.
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In the third row, the required order is available in 2 cases. So we have 2 from the third row.
♦ They are: CABD, CDAB
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In the fourth row, the required order is available in 2 cases. So we have 2 from the fourth row.
♦ They are: DABC, DCAB
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So total number of favorable outcomes =
2 +0 + 2 + 2
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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:
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Number combinations of 7 cards is $\rm{{}^{52} C_{7}}$.
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So we can write: $\rm{n(S) = {}^{52}C_{7}}$
Part (i):
1. We can imagine two boxes.
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The left side box is for the four king cards. The right side box is for the remaining three cards.
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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.
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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.
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The two boxes can be filled together in
$\rm{{}^{4} C_{4}~\times~{}^{48} C_{3}}$ ways.
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Thus the number of combinations with all kings is:
$\rm{{}^{4} C_{4}~\times~{}^{48} C_{3}}$
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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.
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The left side box is for the three king cards. The right side box is for the remaining four cards.
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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.
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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.
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The two boxes can be filled together in
$\rm{{}^{4} C_{3}~\times~{}^{48} C_{4}}$ ways.
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Thus the number of combinations with all kings is:
$\rm{{}^{4} C_{3}~\times~{}^{48} C_{4}}$
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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.
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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)
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We have: P(A∪B) = P(A) + P(B)
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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.
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We can think of a new event (B∪C)
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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”
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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)
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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:
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Five teams can finish in 5! ways.
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So the number of elements in S = 5! = 120
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We can write: n(S) = 120
Part (i):
1. Let G be the event: A, B and C finish first, second and third respectively.
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There are only two possible outcomes for such a finish. They are:
ABC, D, E and ABC, E, D
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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.
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A, B and C can arrange among themselves in 3! ways.
♦ We have, 3! = 3 × 2 = 6
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For each of those 6 ways, D and E can arrange among themselves in 2! ways.
♦ We have, 2! = 2 × 1 = 2 ways.
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So the number of favorable outcomes = 6 × 2 = 12
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We can write: n(H) = 12
2. Since all outcomes in S are equally likely, we get:
$\rm{P(H) = \frac{n(H)}{n(S)} = \frac{12}{120} = \frac{1}{10}}$
Link to a few more solved examples is given below:
In the next section, we will see Appendix A.
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