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 
  S  = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BDAC, BDCA, BCAD, BCDA 
       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. 
  Therefore H = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, CABD, CADB, DABC, DACB} = 12
  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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