An urn contains a number of colored balls, with equal numbers of each color. Adding 20 balls of a new color to the urn would not change the probability of drawing (without replacement) two balls of the same color.

How many balls are in the urn? (Before the extra balls are added.)

n is the number of colors before the 20 balls are introduced

ReplyDeletex is the number of balls in the urn before the 20 balls are introduced

400/(x+20)^2+x^2/(n*(x+20)^2)=1/n

Solve that to get

x=10n-10

So the answer is ten less than ten times the number of colors

There were initially 190 balls in the urn; 10 each of 19 different colors.

ReplyDelete400?

ReplyDelete