Let there be two buckets. Bucket I contains balls of 4 white, 1 red. Bucket II contains balls of 3 white, 5 red and 1 black. Consider the following procedure:
- You draw a ball uniformly at random from Bucket I, put it in Bucket II.
- Then, you draw a ball uniformly at random from Bucket II, put it in Bucket I. The random draw in step 2 is independent of the random draw in step 1.
(a) What is the probability that the resulting Bucket I contains balls of all 3 colours?
(b) What is the probability that the resulting Bucket I contains balls of only 1 colour?
(c) What is the probability that the resulting Bucket I still contains 4 white, and 1 red?
^ For the above questions, I am not sure if I am thinking too much by using conditional probability or if can I just draw out a probability tree to visualise the outcomes.
Let $U$ be the colour of the ball drawn from bucket I, and $V$ be the colour drawn from bucket II, with values in $\{w, r, b\}$ (white, red, and black).
These are just the same task; since the probabilities you assign to branches on the tree are conditional probabilities, and the probabilities you calculate for the leaves are the resultant joint probability
$$\begin{array}{|l:l|l|}\hline\mathsf P(U=w)=4/5&\mathsf P(V=w\mid U=w)=\boxed?&\mathsf P(U=w\cap V=w)=\boxed?\\&\mathsf P(V=r\,\mid U=w)=\boxed?&\mathsf P(U=w\cap V=r)\,=\boxed?\\&\mathsf P(V=b\,\mid U=w)=\boxed?&\mathsf P(U=w\cap V=b)\,=\boxed?\\\hline\mathsf P(U=r)=1/5&\mathsf P(V=w\mid U=r)=\boxed?&\mathsf P(U=r\,\cap V=w)=\boxed?\\&\mathsf P(V=r\,\mid U=r)=\boxed?&\mathsf P(U=r\,\cap V=w)=\boxed?\\&\mathsf P(V=b\,\mid U=r)=\boxed?&\mathsf P(U=r\,\cap V=b)\,=\boxed?\\\hline\end{array}$$
You move a white, or red, ball from bucket I into bucket II, then move a ball back into bucket I. The probabilities of the colour moved outof bucket II will be conditional on the colour moved into it (thus these steps are not independent draws).