Question

There are n counters in a bag.

3 of the counters are red.
Two counters are taken from the bag.
Find, in terms of n, the probability that both counters are red.

Answer 1

`\displaystyle |\Omega|=\binom{n}{2}=(n!)/(2!(n-2)!)=(n(n-1))/(2)\\|A|=\binom{3}{2}=3\\\\P(A)=(3)/((n(n-1))/(2))=(6)/(n(n-1))`