Question 1124483
a) 
1st draw: P(Y) = 3/(5+2+3) = 3/10
2nd draw:  P(G) = 2/(5+2+2) = 2/9 
3rd draw: P(R) = 5/(5+1+2) = 5/8
( each denominator is R+G+Y remaining in the urn )

P(Y, G, R) = (3/10)*(2/9)*(5/8) = 30/720 = {{{ highlight( 1/24 ) }}}<br>

b)
In this case, the number of each color stays fixed (5,2,3 for R,G,Y, respectively), and the denominator obviously stays fixed at 10:

P(Y, G, R) = (3/10)*(2/10)*(5/10) = 30/1000 = {{{ highlight(3/100) }}}<br>
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EDITED 9/27:  Fixed the with-replacement calculation.  I had 3*2*3/1000 when it should have been 3*2*5/1000.  Sorry.  Corrected version is above.