The conclusion is certainly correct.
A simpler way to see it: we don't care about the green balls at all. We are only interested in the three balls, $B, R_1, R_2$. One of these three must be drawn before the others, so by symmetry the answer is $\frac 13$.
lets model out how transfer from b1 to b2 can take place :
case 1) b1 transfers two blue balls to b2 , prob = 0.1
case 2) b1 transfers two red balls to b2 ,prob = 0.3
case 3) b1 transfers a red and a blue ball to b2 , prob = 0.6
lets model the picking of balls from b2 :
case 1.1) picking different colour balls from b2(4b,1r) after case 1 :
prob = 0.4 * 0.1 = 0.04
case 1.2) picking same colour balls from b2(4b,1r) after case 1 :
prob = 0.6 * 0.1 = 0.06
case 2.1) picking different colour balls from b2(2b,3r) after case 2 :
prob = 0.6 * 0.3 = 0.18
case 2.2) picking same colour balls from b2(2b,3r) after case 2 :
prob = 0.4 * 0.3 = 0.12
case 3.1) picking different colour balls from b2(3b,2r) after case 1 :
prob = 0.6 * 0.6 = 0.36
case 3.2) picking same colour balls from b2(3b,2r) after case 1 :
prob = 0.4 * 0.6 = 0.24
given that the balls selected from b2 are different , cases 1.1 , 2.1 , 3.1 apply .
Out of these cases case 1 and 2 had same color transfer from b1 to b2 , case 3 had different colour transfer from b1 to b2 :
prob ( same color transfer from b1 to b2 | differnt colour selection from b2 )
= ( case 1.1 + case 2.1 ) / (case 1.1 + case 2.1 + case 3.1 )
= (0.04+0.18)/(0.04+0.18+0.36) = 0.38
or if you prefer formulas :
$$P(\frac{same:b1->b2}{diff:b2})*p(diff:b2) = P(\frac{diff:b2}{same:b1->b2})*P(same:b1->b2)$$
$$P(\frac{same:b1->b2}{diff:b2}) = \frac{P(\frac{diff:b2}{same:b1->b2})*P(same:b1->b2)}{P(diff:b2)}$$
$$P(\frac{diff:b2}{same:b1->b2})*P(same:b1->b2) = case 1.1 + case 1.2$$
$$P(diff:b2) = case 1.1 + case 2.1 + case 3.1$$
$$P(\frac{same:b1->b2}{diff:b2}) = \frac{case 1.1 + case 1.2}{case 1.1 + case 2.1 + case 3.1}$$
$$P(\frac{same:b1->b2}{diff:b2}) = \frac{(0.04+0.18)}{(0.04+0.18+0.36)} = 0.38$$
Best Answer
Suppose that instead of picking ten balls, you pick all 100 balls and put them into a row in the order you picked them. Every one of the $100!$ possible orders is equally likely, and $20\cdot99!$ have a blue ball in the 10th position. Therefore the probability is exactly $\frac{1}{5}$.