﻿ Class 11 NCERT Math Solution
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TOPICS
Miscellaneous

Example-1 :-  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 number of arrangements (orders) in which Veena can visit four cities A, B, C, or D is 4!
i.e., 24.Therefore, n (S) = 24.
Since the number of elements in the sample space of the experiment is 24 all of
these outcomes are considered to be equally likely.
A sample space for the experiment is
CABD, CADB, CBDA, CBAD, CDAB, CDBA, DABC, DACB, DBCA, DBAC, DCAB, DCBA}
```
```(i) Let the event ‘she visits A before B’ be denoted by E
Therefore, E = {ABCD, CABD, DABC, ABDC, CADB, DACB ACBD, ACDB, ADBC, CDAB, DCAB, ADCB} = 12
Hence, P(E) = n(E)/n(S) = 12/24 = 1/2
```
```(ii) Let the event ‘Veena visits A before B and B before C’ be denoted by F.
Therefore F = {ABCD, DABC, ABDC, ADBC} = 4
Hence, P(F) = n(F)/n(S) = 4/24 = 1/6
```
```(iii) Let the event ‘Veena visits A first and B last’ be denoted by G.
Therefore F = {ACDB, ADCB} = 2
Hence, P(G) = n(G)/n(S) = 2/24 = 1/12
```
```(iv) Let the event ‘Veena visits A either first or second’ be denoted by H.
Hence, P(H) = n(H)/n(S) = 12/24 = 1/2
```
```(v) Let the event ‘Veena visits A either first or second’ be denoted by I.
Therefore I = {ABCD, ABDC, CABD, CDAB, DABC, DCAB} = 6
Hence, P(I) = n(I)/n(S) = 6/24 = 1/4
```

Example-2 :-  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 :-
```  Total number of possible hands = ⁵²C₇

(i) Number of hands with 4 Kings = ⁴C₄ x ⁴⁸C₃ (other 3 cards must be chosen from the rest 48 cards)
Hence P (a hand will have 4 Kings) = (⁴C₄ x ⁴⁸C₃)/⁵²C₇ = 1/7735

(ii) Number of hands with 3 Kings and 4 non-King cards = ⁴C₃ x ⁴⁸C₄
Therefore P (3 Kings) = (⁴C₃ x ⁴⁸C₄)/⁵²C₇ = 9/1547

(iii) P(atleast 3 King) = P(3 Kings or 4 Kings) = P(3 Kings) + P(4 Kings) = 9/1547 + 1/7735 = 46/7735
```

Example-3 :-  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 :-
```  Consider E = B ∪ C so that
P(A ∪ B ∪ C ) = P(A ∪ E ) = P(A) + P(E) - P(A ∩ E) .......(i)
Now, P(E) = P(B ∪ C) = P(B) + P(C) - P(B ∩ C)  .......(ii)
Also A ∩ E = A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)  [using distribution property of intersection of sets over the union].
Thus, P(A ∩ E) = P(A ∩ B) + P(A ∩ C) - P[(A ∩ B) ∩ P(A ∩ C)] = P(A ∩ B) + P(A ∩ C) - P(A ∩ B ∩ C)  .....(iii)

Using (2) and (3) in (1), we get
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P ( A ∩ B ∩ C)
```

Example-4 :-  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 :-
```  If we consider the sample space consisting of all finishing orders in the first three places,
we will have ⁵P₃, i.e., (5!)/(5 - 3)! = 5 × 4 × 3 = 60 sample points, each with a probability of 1/60.

(a) A, B and C finish first, second and third, respectively.
There is only one finishing order for this, i.e., ABC.
Thus, P(A, B and C finish first, second and third respectively) = 1/60

(b) A, B and C are the first three finishers.
There will be 3! arrangements for A, B and C.
Therefore, the sample points corresponding to this event will be 3! in number.
So, P(A, B and C are first three to finish) = 3!/60 = 6/60 = 1/10
```
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