There are 7 b***s in a jar which have been numbered 0, 2, 3, 4, 6, 7, 9. Two b***s are drawn at random and the order in which the b***s are drawn does not matter. What is the probability of drawing a ball numbered with 2 and a ball numbered with 6?
The possible permuted sets of choosing 2 b***s from 7 = P(7,2) = 42
But the only ones we are interested in are {2,6} or {6,2}
So.....
2 / 42 = 1 / 21 ≈ 4.76%
Thank you CPhill!
OK......no prob......