Latest Gate Aptitude Question SOLUTION: 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

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.

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

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