Conditional Probability using an urn. An urn contains $5$ white balls and $6$ black balls. $3$ balls are drawn sequentially without replacement. What are the probabilities...
a. $P(\text{2nd is black} | \text{1st is white})$?
For this one I did the
$$ P(\text{1st is white})=\frac{5}{11} , $$
and
$$ P(\text{second is black}) = \frac{3}{11}*\frac{3}{11} $$
(that is the probability of WB and BB added together). I multiplied the probability of
$$ \frac{P(\text{2nd black} \cap \text{1st white})}{P(\text{1st is white})} , $$
for this I got
$$ \frac{\frac{5}{11}*\frac{3}{11}}{\frac{5}{11}} = \frac{3}{11} . $$
However since this is without replacement I did not think that the answer could be independent, some feedback would help. Alright so I have this figured out that it should be $\frac{6}{10}$.
b. If I am looking now for $P(\text{3rd is black} | \text{1st is white and 2nd is black})$, I ended up with
$$ \frac{\frac{5}{9}*\frac{3}{11}}{\frac{3}{11}} , $$
this is again an independent answer is this correct?
 A: There's no need to use any formulas here, just look at what happens when you follow the conditions and remove the relevant balls from the urn step by step.
If we start with $5$ white balls and $6$ black balls, then we're given that the first ball drawn is white, then after that first draw, $6$ black balls remain and $5-1=4$ white balls remain. So, we have $6$ black balls out of $10$ total balls, and the answer to the first question is
$$ P[\text{2nd is black} | \text{1st is white}] = \frac{6}{10} . $$
For the second question, we're given that the first is white and the second is black. So, after these two draws, we know that $5-1=4$ white balls remain and $6-1=5$ black balls remain. Out of a total of $9$ remaining balls, $5$ are black, so we have
$$ P[\text{3rd is black} | \text{1st is white and 2nd is black}] = \frac{5}{9} . $$
A: Using the notation $B_n$ to represent the events "the $n$-th ball is black", and $W_n$ likewise for white balls.
$$\mathsf P(W_1)=\dfrac{5}{11}\\\mathsf P(B_2)=\dfrac 6{11}$$
Reason: each individual ball has the same chance to be the second one drawn there are six black balls among the eleven.

(that is the probability of WB and BB added together).

That is true, but for some reason you multiplied.  You should add.
$$\begin{align}\mathsf P(B_2)&=\mathsf P(W_1\cap B_2)+\mathsf P(B_1\cap B_2)\\[1ex] &=\mathsf P(W_1)\cdot\mathsf P(B_2\mid W_1)+\mathsf P(B_1)\cdot\mathsf P(B_2\mid B_1)\\[1ex]&=\dfrac{5}{11}\dfrac{6}{10}+\dfrac{6}{11}\dfrac{5}{10}\\[1ex]&=\dfrac 6{11}\end{align}$$
Here, $\mathsf P(B_2\mid W_1)$ is evaluated as the probability for drawing one from the $6$ black balls when selecting one from the $10$ balls remaining after one white ball has been removed.


I multiplied the probability of
$$ \frac{P(\text{2nd black} \cap \text{1st white})}{P(\text{1st is white})} = \frac{\frac{5}{11}*\frac{3}{11}}{\frac{5}{11}} \\= \frac{3}{11} .
 $$

You had already found  $\mathsf P(\text{1st white$\cap$ 2nd black})$ to be just $3/11$.  So you wanted:$$\mathsf P(B_2\mid W_1)=\dfrac{\mathsf P(W_1\cap B_2)}{\mathsf P(W_1)}=\dfrac{3/11}{5/11}=\dfrac{3}{5}$$
Which, again, you had already evaluated without recognising it.
A: I'm not sure what you did, but
P(2 is B|1 is W) = P(pick B among 4W6B) = 6/10
A: a. P(second is black|first is white)=P(black ball out of 4 white 6 black)=6/10
b. P(third is black|1st is white and 2nd is black)=P(black ball out of 4 white and 5 black)=5/9
