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  • Vijaykumar Nayak (RBIN/EDS1) *
    1.A box has W white marbles and B black marbles ... show that probability of picking a white marble in the first try =probability of picking a white marble in
    Message 1 of 5 , Nov 4, 2003
      1.A box has W white marbles and B black marbles ...
       
      show that
      probability of picking a white marble in the first try =probability of picking a white marble in second try with first marble already picked
       
      or in other words, Prove that
       
      P(W1) = P(W2) =  W/(W+B)
       
       
       
    • RajSekhar
      hi, good to see some puzzle here.....keep going on posting the puzzles...... P(W1) = W/(W+B).......this is obvious now,for P(W2) = P1 + P2 where P1 =
      Message 2 of 5 , Nov 4, 2003
        hi,
         
        good to see some puzzle here.....keep going on posting the puzzles......
         
        P(W1) = W/(W+B).......this is obvious
         
        now,for P(W2) = P1 + P2
        where P1 = probability that the first ball is white and second ball is also white
        P2 = probability that the first ball is black and second ball is white
         
        P1 =  W/(W + B)   x  ( W - 1)/(W + B - 1)
         
        P2 = B/(W + B)   x  W/(W + B - 1)
         
        adding, P1 + P2 = (W^2 - W + WB)/(W + B)x(W + B -1) = W/(W + B)
         
         
         
         
         
        raj 
        "Vijaykumar Nayak (RBIN/EDS1) *" <Vijaykumar.Nayak@...> wrote:
        1.A box has W white marbles and B black marbles ...
         
        show that
        probability of picking a white marble in the first try =probability of picking a white marble in second try with first marble already picked
         
        or in other words, Prove that
         
        P(W1) = P(W2) =  W/(W+B)
         
         
         


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      • Ankur Jain
        P(W1) = W/(W+B) P(W2) = [W/(W+B) * (W-1)/(W+B-1)] + [B/(W+B) * W/(W+B-1)] = [W(W-1) + B*W] / [(W+B)*(W+B-1)] = [W(W+B-1)] / [(W+B)*(W+B-1)] = W/(W+B)
        Message 3 of 5 , Nov 4, 2003
          P(W1) = W/(W+B)

          P(W2) = [W/(W+B) * (W-1)/(W+B-1)] + [B/(W+B) * W/(W+B-1)]

          = [W(W-1) + B*W] / [(W+B)*(W+B-1)]

          = [W(W+B-1)] / [(W+B)*(W+B-1)]

          = W/(W+B)

          > 1.A box has W white marbles and B black marbles ... show that
          > probability of picking a white marble in the first try =probability of
          > picking a white marble in second try with first marble already picked
          > or in other words, Prove that
          > P(W1) = P(W2) = W/(W+B)
        • Atul Mathur
          P(w1) is obviously W/(W+B), since he has to choose from W+B marbles out of which W are white.. Now P(w2) = probability of getting a white marble in the second
          Message 4 of 5 , Nov 6, 2003
            P(w1) is obviously W/(W+B), since he has to choose
            from W+B marbles out of which W are white..

            Now P(w2) = probability of getting a white marble in
            the second turn with a white one having been chosen in
            the first + probability of a white marble being chosen
            with a black marble in the first chance

            = (W/(W+B))*((W-1)/(W+B-1)) + (B/(W+B))*((W)/(W+B-1))

            = ((W(W-1) + BW))/((W+B-1)(W+B))

            = W(W+B-1)/((W+B-1)(W+B))

            = W/(W+B)


            Hope this satisfies you..

            Bye,
            Atul.


            --- "Vijaykumar Nayak (RBIN/EDS1) *"
            <Vijaykumar.Nayak@...> wrote: > 1.A box has W
            white marbles and B black marbles ...
            >
            > show that
            > probability of picking a white marble in the first
            > try =probability of picking a white marble in second
            > try with first marble already picked
            >
            > or in other words, Prove that
            >
            > P(W1) = P(W2) = W/(W+B)
            >
            >
            >
            >

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          • kandey kranthi kumar
            Probability of picking a white ball in first try is W/(W+B). Now comming to picking a white ball in second try..here two cases picking a while ball in first
            Message 5 of 5 , Nov 10, 2003

              Probability of picking a white ball in first try is W/(W+B).

              Now comming to picking a white ball in second try..here two cases
              picking a while ball in first try or picking a black ball in first try.

              case 1:
              Probabilityy of picking a white ball in first try and white ball in second try is
              W/(W+B) X (W-1)/(W+B-1) = W(W-1)/(W+B-1)(W+B)

              Case 2:
              Probability of picking a black ball in first try and white ball in second try is
              B/(W+B) X (W)/(W+B-1) = BW/(W+B-1)(W+B)

              So total probability is addition of above two
              which is (BW + W(W-1))/(W+B-1)(W+B) = W(W+B-1)/(W+B-1)(W+B) = W/(W+B)
              Hence proved

              -kranthi

              On Tue, 04 Nov 2003 Vijaykumar Nayak (RBIN/EDS1) * wrote :

              >1.A box has W white marbles and B black marbles ...
              >
              >show that
              >probability of picking a white marble in the first try =probability of picking a white marble in second try with first marble already picked
              >
              >or in other words, Prove that
              >
              >P(W1) = P(W2) =  W/(W+B)
              >
              >
              >



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