Question 1038041
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?
<pre>
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) = {{{(4/16)(6/15)(4/14)}}} = {{{(1/4)(2/5)(2/7)}}} = {{{1/35}}}

So answer = {{{6*(1/35)}}} = {{{6/35}}}

Edwin</pre>