Re: Daily question

Posted by Khushboo on Feb 25, 2009; 2:04pm
URL: http://discussion-forum.276.s1.nabble.com/Daily-question-tp2328964p2383864.html

1(a) P(B|A)+P(B|A') = 1, for example
     A coin is tossed twice. Let A and B be d events given by
     A - the first toss is a tail
     B - there is a head on second toss
    n(s) = (H,H) , (H,T) , (T,H) , (T,T)
   so,
      P(B/A) = P(A ∩  B) / P(A) = 1/4 / 2/4 = 1/2
      P(B/A') = 1/4 / 2/4 = 1/2
   therefore,
      P(B|A)+P(B|A') = 1/2 + 1/2 = 1

 (b) P(B|A)+P(B|A')   not be equal to 1, for example
      Suppose a bag contains 5 white and 4 red balls. Two balls r drawn from d bag one after the other without replacement. Let A and B be d events given by
     A - Drawing a white ball in d 1st draw.
     B - Drawing a red ball in d 2nd draw.
 Now,
    P(B/A) = prob of drawing a red ball in 2nd draw given dat a white ball has already been drawn in d 1st draw.

    Since 8 balls r left after drawing a white ball in 1st draw and out of these 8 balls , 5 balls r red.
  so,
  P(B/A) = 4/8
  P(B/A') = 3/8

 therefore, P(B|A)+P(B|A') = 4/8 + 3/8 = 7/8 = 0.875


2- Given three events A, B and C such that P(A ∩ B ∩ C) ≠ 0, and P(C|A ∩ B) = P(C|B), show that P(A|B ∩ C) = P(A|B)

   We knw dat, P(C|A ∩ B) = P(C|B)
                     P(C∩A∩B) / P(A∩B) = P(C∩B) / P(B)
                     P(A∩B∩C) = P(A∩B) P(B∩C) / P(B)
   therefore,
                P(A|B ∩ C) = P(A∩B∩C) / P(B∩C)
                               = P(A∩B) P(B∩C) / P(B)  /  P(B∩C)
                               = P(A∩B) / P(B)
                               = P(A|B)
                                  Hence prooved