Probability of removing at least 1 white ball given randomly selected remaining ball is white Followup to this question: Probability of selecting 1 white ball given $t$ removed balls
Let there be an urn of $b$ balls, $w$ of which are white. Uniformly and randomly throw away exactly $t,(0≤t<b)$ balls. Then randomly select one of the remaining $b−t$ balls. 
Given that the randomly selected ball is white, what is the probability that at least one of the thrown away balls is white?
I would appreciate some feedback on my work as I'm not sure what I'm doing wrong. So if we let $A$ be the event that that the randomly selected ball is white and $B$ be the event that at least one of the thrown away balls is white. Then we want to find $P(B|A) = \frac{P(A\cap B)}{P(A)}$
Since $A, B$ aren't necessarily independent events, then we must solve $P(A \cap B) = P(B)*P(A|B)$.
I know that $P(B) = 1 - \frac{{b-w \choose t}}{b \choose t}$ from limiting the range of PMF of the problem to $[1,c]$ instead of $[0,c]$.
I also know that $P(A) = \frac{w}{b}$.
But when I solve for $P(A|B)$ using PMF and solution presented in the link above, I just get $P(A)*P(B)$ which doesn't make a lot of sense to me. Please advise. Thank you.
Also feedback on my solution would be nice too
Edit: Typo
 A: Here is a third answer:


*

*$P(A)$ is the probability that the $t+1$th randomly selected ball is white

*$P(B)$ is the probability that at least one of the $t$ thrown away balls are white

*$P(B^c)=1-P(B)$ is the probability that none of the $t$ thrown away balls are white
Then I would have thought 


*

*$P(A)=\dfrac{w}{b}$

*$P(B^c) = \dfrac{{b-w \choose t}}{b \choose t}$

*$P(A \mid B^c) = \dfrac{w}{b-t}$

*$P(A \cap B^c) = \dfrac{w}{b-t}\dfrac{{b-w \choose t}}{b \choose t}$

*$P(B^c \mid A) = \dfrac{b}{b-t}\dfrac{{b-w \choose t}}{b \choose t}=\dfrac{{b-w \choose t}}{b-1 \choose t}$

*$P(B \mid A) = 1 - \dfrac{{b-w \choose t}}{b-1 \choose t}$
As an illustration of the differences, consider $w=2,b=4,t=2$: my expression gives an answer of $\frac23$ while molloculle's answer gives $\frac{5}{12}$ and  Graham Kemp's gives $\frac12$.  I think that if the third ball is white, there is a $\frac23$ probability the other white ball was one of the first two thrown away and a $\frac13$ probability the other white ball was the remaining fourth ball.  Looking at it another way in this example, of the $4!=24$ ways of ordering the four balls, $12$ have the third ball white and $8$ of those $12$ have at least one of the first two balls white, with $\frac{8}{12}=\frac23$ 
A: You have a heap of $b$ balls, of which $w$ are white. You are randomly arranging that into three heaps of $t$ discarded, $1$ selected, and $b-t-1$ remaining balls; with $t$ uniformly distributed over $\{0..b-1\}$.   You seek the probability for not-none of the discarded balls are white given that the selected ball is.
For a particular $t$, there are $\binom{b-t}{w}$ ways that none of the white balls are among the discards.   There are $\binom b w$ ways to arrange the balls in total.   Now $t$ varies between $0$ to $b-w$ (as any further and it is impossible to not-discard some white balls), with each value occuring with probability of $1/b$.
Hence $$\begin{align}\mathsf P(B)&=1-\dfrac 1b\sum\limits_{t=0}^{b-w}\dfrac{\dbinom{b-t}w}{\dbinom bw}\\[1ex]&~\vdots\tag{why?}\\[1ex] &=1-\dfrac 1b\cdot\dfrac{(b+1)}{(w+1)}\\[1ex] &=\dfrac{bw-1}{bw+b}\end{align}$$

There are $\binom{b-t-1}{w-1}\binom 11$ ways none of the white balls are among the discards and the selected ball is white.   There are $\binom{b-1}{w-1}\binom 11$ ways the selected ball may be white, in total.   Also $t$ behaves as before.
$$\mathsf P(B\mid A)=1-\dfrac 1b\cdot\dfrac{(b-1)+1}{(w-1)+1)}=\dfrac{w-1}w$$
A: I believe the correct answer for this should be:
$$\frac{n-c-p}{n-p}*(1 - \frac{{b-w \choose t}}{{b \choose t}})*\frac{n}{c}$$
I did this a slightly different method than how Mr. Kemp did it. 
Let $P(A)$ represent the probability that our randomly selected ball is white
Let $P(B)$ represent the probability that at least one of the thrown away balls are white
Since we know that 
$$P ( B | A ) = \frac { P ( A \cap B ) } { P ( A ) }$$
And $A, B$ are independent events, so we must substitute $P ( A \cap B ) = P ( B ) * P ( A | B )$, we can find each of these. 
$P(A) = \frac{w}{b}$ from the linked problem.
$P(B) = 1 - \frac{{b-w \choose t}}{{b \choose t}}$ 
I found this by limiting the following sum:
$$\sum _ { v } \frac { \left( \begin{array} { c } { w } \\ { v } \end{array} \right) \left( \begin{array} { l } { b - w } \\ { t - v } \end{array} \right) } { \left( \begin{array} { l } { b } \\ { t } \end{array} \right) }$$
from $v = [1, n]$ instead of $[0,c]$, that is, I subtracted $v = 0$ from the sum. 
$P(A|B) = \frac{n - c - p}{n - p}$ We can find this by complementary counting. If we instead find the probability of $A$ given $\overline{B}$, that is the event that we throw away no white balls, then that probability is $\frac{c}{n-p}$. We then take $1 - \frac{c}{n-p}$.
Now substitute everything back into the conditional probability equation and we get that 
$$\frac{n-c-p}{n-p}*(1 - \frac{{b-w \choose t}}{{b \choose t}})*\frac{n}{c}$$
