There is an urn with three red balls, three blue balls, and three green balls. Suppose three balls are randomly drawn from the nine balls, and then three more balls will be randomly drawn from the six that were not selected on the first draw. What is the probability that the balls in the first subset drawn will all be the same color, and that the balls in the second subset drawn will all be the same color?
Attempted Solution:
So the first ball can be any color, but then the next two would have to be the same color as the first ball so we have,
$\frac{9*2*1}{9*8*7}$
Then the fourth ball can be any of the remaining six, but the next two balls will have to be the same color as the fourth ball so we have,
$\frac{9*2*1*6*2*1}{9*8*7*6*5*4}$ = $.003571$
Is this a valid solution?