# Probibility extended response question

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.

• Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. Commented Oct 23, 2019 at 6:36
• A) Use law of total probability and condition on the outcome of the draw from the red box. B) use Bayes theorem Commented Oct 23, 2019 at 6:37

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).