I've been prepping for my end of year exams and I'm having difficulties answering this particular question. I don't recall the rules for these types of questions, any help with the solving methodology would be appreciated. Thanks!
There are n identical black balls and n identical white balls. A blue box contains 3 black balls and n−3 white balls. A red box contains n−3 black balls and 3 white balls. A ball is taken at random from the red box and put in the blue box. A ball is then taken at random from the blue box.
(A) Find the probability, in terms of n, that the ball taken from the blue box is:
i) black
ii) White
(B) Find the probability, in terms of n , that the first ball is black given that the second is white.
We will use the phrases "$1^{\text{st}}$" and "$2^{\text{nd}}$" to mean, respectively, the ball being drawn from the red box (then put in the blue box) and the ball subsequently drawn from the blue box. [This is consistent with the phrasing of part (B) in the question.] Also, we'll use the obvious labels "$\text{B}$" and "$\text{W}$" for black and white.
(A) There are two possibilities: either $1^{\text{st}}$ is $\text{B}$, or $1^{\text{st}}$ is $\text{W}$. The probability that $1^{\text{st}}$ is $\text{B}$ is $\,\frac{n-3}{n},\,$ and in that case the probability that $2^{\text{nd}}$ is $\text{B}$ is $\,\frac{4}{n+1},\,$ and the probability that $2^{\text{nd}}$ is $\text{W}$ is $\,\frac{n-3}{n+1}.\,$ The computations will be similar for the cases when $1^{\text{st}}$ is $\text{W}$.
Then for (i) we have that
$$ \color{white}{text}\\ \begin{align} P\;[2^{\text{nd}} \text{is B}]&=P\;[1^{\text{st}} \text{is B}\; \land \; 2^{\text{nd}} \text{ is B}] + P\;[1^{\text{st}} \text{is W} \land 2^{\text{nd}} \text{ is B}]\\[2ex] &=\left(\frac{n-3}{n}\right)\left(\frac{4}{n+1}\right)+ \quad...\\ \color{white}{text} \end{align} $$
Can you complete it from there? Part (ii) should be quite similar.
(B) This is asking for $\,P\;[1^{\text{st}} \text{is B} \mid 2^{\text{nd}} \text{ is W}].$ Using the formula $\,P\;[B \mid A] = \frac{P\;[B \land A]}{P\;[A]},\,$ we get
$$ \color{white}{text}\\ \begin{align} P\;[1^{\text{st}} \text{is B} \mid 2^{\text{nd}} \text{ is W}] &= \frac{P\;[1^{\text{st}} \text{is B} \land 2^{\text{nd}} \text{ is W}]}{P\;[2^{\text{nd}} \text{ is W}]}\\[2ex] &= \quad ...\\[2ex] \color{white}{text}\\ \end{align} $$
where both the numerator and denominator would already have been computed in part (A).