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 One bead is randomly withdrawn from each bag.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
 A: 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 $P(X\ge2)$.
A: 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
$${3\choose2}\cdot \left({3\over10}\right)^2{7\over10}+{3\choose 3}\left({3\over10}\right)^2={27\over125}=0.216\ .$$
A: 
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 


No that the probability of not drawing a white bead with a black mark given that you so marked only two of the three white beads in each bag.
"At least two" is "not one nor none".   That is, if we let $W$ be the count of white marbles drawn: $$\begin{align}\mathsf P(W\geq 2)~=~&1-\mathsf P(W=0)-\mathsf P(W=1)\\ =~& 1 - (\tfrac 7{10})^3 - 3(\tfrac 7{10})^2\tfrac 3{10} \\ =~& \tfrac{1000-343-441}{1000} \\ =~& \tfrac{216}{1000} \\ =~& \dfrac {27}{125}\end{align}$$
$\mathsf P(W=0)=\tfrac 7{10}\tfrac 7{10}\tfrac 7{10}$ as it is the probability of obtaining not-white from each bag.
$\mathsf P(W=1)= \tfrac 3{10}\tfrac 7{10}\tfrac 7{10}+\tfrac 7{10}\tfrac 3{10}\tfrac 7{10}+\tfrac 7{10}\tfrac 7{10}\tfrac 3{10} =3(\tfrac 7{10})^2\tfrac 3{10}$ as it is the probability of obtaining not-white from two bags and white from the other in any of the $3$ orders the white may emerge.
Alternatively
$$\begin{align}\mathsf P(W\geq 2)~=~&\mathsf P(W=2)+\mathsf P(W=3)\\ ~=~& 3(\tfrac 3{10})^2\tfrac 7{10} + (\tfrac 3{10})^3 \\ ~=~& \dfrac {27}{125}\end{align}$$
