mathschallenge.net logo

Equal Chance

Problem

A bag contains n discs, made up of red and blue colours. Two discs are removed from the bag.

If the probability of selecting two discs of the same colour is 1/2, what can you say about the number of discs in the bag?


Solution

Let there be r red discs, so P(RB) = r/n times (nminusr)/(nminus1), similarly,
P(BR) = (nminusr)/n times r/(nminus1).

Therefore, P(different) = 2r(nminusr)/(n(nminus1)) = 1/2.

Giving the quadratic, 4r2 minus 4nr + n2 minus n = 0.

Solving, r = (nplus or minusradicaln)/2.

If n is an odd square, radicaln will be odd, and similarly, when n is an even square, radicaln will be even. Hence their sum/difference will be even, and divisible by 2.

In other words, n being a perfect square is both a sufficient and necessary condition for r to be integer and the probability of the discs being the same colour to be 1/2.

Prove that n(n+1)/2 (a triangle number), must be square, for the probability of the discs being the same colour to be 3/4, and find the smallest n for which this is true.
What does this tell us about n and n(n+1)/2 both being square?
Can you prove this result directly?

Problem ID: 146 (Jan 2004)     Difficulty: 3 Star

Only Show Problem