In ram's house there was a basket full of balls.The basket has a number of balls which are coloured, with equal numbers of each color. One day, his uncle shyam came and gave him 20 balls of a new colour. Adding 20 balls of a new color to the basket would not change the probability of drawing (without replacement) two balls of the same color.

How many balls are in the urn originally?

• We now calculate the probability of drawing matching colors, before and after adding the extra balls. All drawings are understood to be without replacement.
  
  Before
  
  Let there initially be cn balls; comprised of c colors, with n > 1 balls of each color.
  The number of ways of drawing two balls is cn(cn − 1). (There are cn choices for the first ball; cn − 1 choices for the second.)
  The number of ways of drawing two balls of a particular color is n(n − 1).
  Summing over all colors, the number of ways of drawing matching colors is cn(n − 1).
  
  Hence the probability of drawing matching colors is
  
  After
  
  Let k balls of a new color be added. (We will set k = 20 at an appropriate point.)
  The number of ways of drawing two balls is (cn + k)(cn + k − 1).
  The number of ways of drawing matching colors is cn(n − 1) + k(k − 1).
  
  Hence the probability of drawing matching colors is
  
  Equating Before and After
  
  Equating the above before and after probabilities, we get
  
  (cn − 1)[cn(n − 1) + k(k − 1)] = (n − 1)(cn + k)(cn + k − 1)
  
  Expanding, we have
  
  c^2n^3 − cn^2 − c^2n^2 + cn + cnk^2 − cnk − k^2 + k = c^2n^3 + 2cn^2k + nk^2 − cn^2 − nk − c^2n^2 − 2cnk − k^2 + cn + k
  
  Simplifying, we find that most terms cancel, yielding
  
  cnk^2 = 2c(n^2)*k + nk^2 − nk − cnk
  
  Dividing by nk (which is non-zero), and regrouping
  
  c(k + 1 − 2n) = k − 1
  
  Substituting k = 20, we get c(21 − 2n) = 19.
  
  The only solution with c > 1 is c = 19, n = 10.
  
  Hence there were initially 19 × 10 = 190 balls in the urn.
• 11 years agoHelpfull: Yes(5) No(0)
• let initially no. of balls=x
  as given
  probability of drawing two balls is also same
  then
  xC2=(x+20)C2
  on solving,x=90
• 11 years agoHelpfull: Yes(2) No(3)
• Answer 190,10 balls of 19 colours.
  
  suppose there are x balls of n colours,then we have
  
  x-1/nx-1=(nx(x-1)+380)/(nx+20)(nx+19)
  
  (x-1)(40nx+380)=380(nx-1)
  .
  .
  
  nx=(21n-19)/2
  
  now n=19 we get nx=190
  
  
  
• 11 years agoHelpfull: Yes(1) No(3)