A bag contains 4 red, 6 blue, 4 yellow, and 2 green marbles.
Once a marble is selected it is not replaced. What is the
probability that in 3 successive draws you get exactly one blue,
one red and no green?
That means you MUST get exactly 1 red, exactly 1 blue, and
exactly 1 yellow. Let R&B&Y mean that you draw a red first,
a blue second, and a yellow third. Since it does not matter
what order they are drawn in, the desired probability is:
P(R&B&Y)+P(R&Y&B)+P(B&R&Y)+P(B&Y&R)+P(Y&R&B)+P(Y&B&R)
Each of those 3!=6 orders in which they could be drawn
has the same probability:
P(R&B&Y) = = =
So answer = =
Edwin