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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 ($n$minus$r$)/($n$minus1), similarly,
P(BR) = ($n$minus$r$)/$n$ times $r$/($n$minus1).

Therefore, P(different) = 2$r$($n$minus$r$)/($n$($n$minus1)) = 1/2.

Giving the quadratic, 4$r$2 minus 4$nr$ + $n$2 minus $n$ = 0.

Solving, $r$ = ($n$plus or minusradical$n$)/2.

If $n$ is an odd square, radical$n$ will be odd, and similarly, when $n$ is an even square, radical$n$ 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?

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

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