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 One bead is randomly withdrawn from each bag.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

We are drawing one bead each from the three bags, and we are drawing them independently. There is a $\frac3{10}$ chance of drawing a white bead from each bag. If we get at least two white beads, we could have got them from


*

*the nylon and plastic bags

*the leather and nylon bags

*the leather and plastic bags

*all three bags


For each of the first three cases it is implied that we draw a non-white bead from the third bag, which has a $\frac7{10}$ chance of happening. The probability of each of these cases happening is therefore $\frac3{10}\times\frac3{10}\times\frac7{10}=\frac{63}{1000}$; we multiply this by three for the probability of getting exactly two white beads, which works out to be $\frac{189}{1000}$.
Similarly, the last case (drawing exactly three white beads) has probability $\frac3{10}\times\frac3{10}\times\frac3{10}=\frac{27}{1000}$ of occurring. Since drawing two white beads and drawing three white beads are mutually exclusive events, add them together to get your answer: $\frac{189}{1000}+\frac{27}{1000}=\frac{216}{1000}=\frac{27}{125}$.
If you look a little deeper this is really a binomial distribution $X$ with $n=3$ and $p=\frac3{10}$; we have just calculated $P(X\ge2)$.
