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A urn contains four red and six white balls. X performs two trials in a sequence as follows:

In the first trial, X picks a ball from the urn at random and marks its colour. If X gets a red ball, then X returns two red balls (i.e., with one additional red ball) back into the urn. If X gets a white ball, X returns three white balls (i.e., with two additional white balls) back into the urn.In the second trial, X randomly selects a ball from that urn again and marks its colour.

How do I calculate the probability that X picked two balls with different colours from the two trials? Should I apply Bayes theorem ? I'm new to probability.

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    $\begingroup$ The experiment is so small and simple that you can scarcely do better than enumerate all possible outcomes and compute their probabilities. There are only four possibilities for the two draws. $\endgroup$
    – whuber
    Apr 20, 2021 at 15:13

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As @whuber said, there are so few possibilities here that a probability tree (comprehensive enumeration) is the simplest approach.

For the first draw at time $t = 1$, there are two possibilities: $R$ and $W$. $P(R) = 0.4$ and $P(W) = 0.6$. Here comes the tree:

IF a red was chosen at $t = 1$, THEN two reds are returned (one is added to the existing ten) and ball is drawn. Now at $t=2$ $P(R) = \frac{5}{11}$ and $P(W) = \frac{6}{11}$.

IF a white was chosen at $t = 1$, THEN three whites are returned (two are added to the existing ten) and ball is drawn. Now at $t=2$ $P(R) = \frac{4}{12}$ and $P(W) = \frac{8}{12}$.

So the probability of picking two different colored balls in the two draws would be red then white or white then red. Given the exhaustive list above, we can see that: $$ \begin{aligned} P(RW) = \frac{4}{10}\cdot\frac{6}{11} = \frac{12}{55}\\ P(WR) = \frac{6}{10}\cdot\frac{4}{12} = \frac{11}{55}\\ \rule{4cm}{0.3pt}\\ P(RR) = \frac{4}{10}\cdot\frac{5}{11} = \frac{10}{55}\\ P(WW) = \frac{6}{10}\cdot\frac{8}{12} =\frac{22}{55} \end{aligned} $$

So the probability of getting two different colors is: $$ \frac{12}{55} + \frac{11}{55} = \mathbf{\frac{23}{55} \approx 41.8\%}. $$

For completeness, it is good to see that the probability of getting two of the same color is $\frac{32}{55} \approx 58.2\%$ and that all the probabilities sum to 1, as they should.

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