How do I solve a problem consisting of independent events? Question:
A leather bag contains 4 black beads, 3 red beads and three white beads. Inside a plastic bag are 5 black beads, 2 red beads and 3 white beads. Another nylon bag contains 6 black beads, 1 red bead and 3 white beads. One bead is randomly withdrawn from each bag. 
What is the probability of getting at least two white beads?
My workout:

1- (4/5 * 4/5 * 4/5) # the probability of not getting 2 white beads and the only way for that to happen is if only 1 white bead is picked therefore leaving 8 to pick from. And since each bag has 10 beads including 3 white beads the chance for this is 8/10= 4/5 This is my logic behind this 

My final answer:

1- (4/5 * 4/5 * 4/5) = 1 - 64/125 = 61/125

My problem with this now is, I don't know if this is the answer or I have missed a step. Any help will be appreciated

There are three independent trials with success probability ${3\over10}$  for each. In such a case the number of successes is binomially distributed. The probability of at least two successes then comes to
$${3\choose2}\cdot \left({3\over10}\right)^2{7\over10}+{3\choose 3}\left({3\over10}\right)^2={27\over125}=0.216\ .$$
