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?
Let there be $r$ red discs, so P(RB) = $r$/$n$ ($n$$r$)/($n$1), similarly,
P(BR) = ($n$$r$)/$n$ $r$/($n$1).
Therefore, P(different) = 2$r$($n$$r$)/($n$($n$1)) = 1/2.
Giving the quadratic, 4$r$2 4$nr$ + $n$2 $n$ = 0.
Solving, $r$ = ($n$$n$)/2.
If $n$ is an odd square, $n$ will be odd, and similarly, when $n$ is an even square, $n$ 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?