Question 1062407
This is a Combinations problem.
Number of ways of choosing any 2: 8C2 = 8!/((8-2)!2!) = 28. This will be the denominator
P[0] = 3C0*5C2/28 = ((3!/((3-0)!0!)*(5!/((5-2)!2)))/28 5/14
OK, here are the other two Probabilities, you can do the details like I did, just copy my work:
P[1] = 3c1*5c1/28 = 15/28
P[2] = 3c2*5c0/28 = 3/28