probability with replacement A box contains 4 red, 3 black and 2 white cubes.  A cube is randomly drawn and has its color noted.  The cube is then replaced, together with 2 more of the same color.  A second cube is then drawn.  Find the probability that 
1.  The second cube is black
2.  the second cube is not white

Divide into cases. With probability $\frac{4}{9}$ a red is drawn. If that happens, the probability of a black next is $\frac{3}{11}$.
With probability $\frac{3}{9}$ a black is drawn. If that happens, the probability of a black next is $\frac{5}{11}$.
With probability $\frac{2}{9}$ a white is drawn. If that happens, the probability of a black next is $\frac{3}{11}$.
Thus the probability that the second ball drawn is black is $\frac{4}{9}\cdot\frac{3}{11}+\frac{3}{9}\cdot\frac{5}{11}+\frac{2}{9}\cdot\frac{3}{11}$. 
We could have saved some work on this part of the question by noting that the only thing that is relevant is whether the first ball drawn is black or not. If the first ball drawn is not black, the probability of a black next is $\frac{3}{11}$. If the first ball drawn is black, the probability of a ball next is $\frac{5}{11}$. That gives answer $\frac{6}{9}\cdot \frac{3}{11}+\frac{3}{9}\cdot\frac{5}{11}$. 
The second problem is done in basically the same way, and is left to you. 
