Drawing three marbles from a box with six marbles, two of which are black 
A box contains six marbles, two of which are black. Three are drawn with replacement. What is the probability two of the three are black?

My attempt:
$$P=\frac{\binom22\binom41}{\binom63}=\frac15$$
However, I think my answer is true when there is no replacement. How should I approach this question when there is replacement?
 A: 
A box contains six marbles, two of which are black. Three are drawn with replacement. What is the probability two of the three are black?

There are $6^3=512$ ways to draw three marbles. The following are all the favorable outcomes where B = Black and X = other. They are:
$$B,B,X,= \frac2{27}$$$$ B,X,B,= \frac2{27}$$$$ X,B,B= \frac2{27}$$ 
Adding all of these up, the answer is $\frac29$.
A: Sampling from the box with replacement will return a black marble $\frac13$ of the time. In the three samples you draw, if two are to be black, the non-black marble could be drawn first, second or third – three possibilities in total. For each case the probability of it happening is
$$\frac13\times\frac13\times\frac23=\frac2{27}$$
where the first two terms give the probability of drawing a black marble, and the last gives that of drawing a non-black marble. Multiplying this by 3 we get a final probability of $\frac2{27}\times3=\frac29$ for drawing two black marbles in our three samples with replacement.
A: I think replacement is easier.  
You have a $\frac {1}{3}$ chance that any marble on any selection is black. 
$(\frac {1}{3})^2$ the first 2 are black. 
$(\frac {1}{3})^2\frac 23$ to pick BBW in that order, 
And, there are 3 combinations of order.  
$\frac {2}{9}$ of getting exactly 2.
the more cut-to-the-chase approach is to consider the $2^{nd}$ term of the binomial expansion of $(a+b)^3$ where $a = \frac 13$ and $b = \frac 23$  
$\frac {7}{27}$ of getting at least 2 (exactly 2 or all 3).
A: 
However, I think my answer is true when there is no replacement. How should I approach with replacement?

Yes, your answer is correct for extraction without replacement.   That is for a hypergeometric distribution; in this case the probability for drawing two successes in a sample of three items from a population of six items containing two successes.
When drawing with replacement, each draw becomes independent of the others and has identical rate of success.   The count of successes in a sequence of such draws has a binomial distribution.
The success rate of each draw is $2/6$.   You want the probability for attaining two successes and one failure in any order.  $$\binom{3}{2}~{\big(\tfrac 13\big)}^2{\big(\tfrac 23\big)}^1~=~\dfrac{2}{9}$$
