# There are three red and two yellow balls inside an urn. Players A and B select balls from the urn without replacement until a yellow ball is selected.

There are three red and two yellow balls inside an urn. Players A and B select balls from the urn without replacement until a yellow ball is selected. Assume that player A selects first. Find the probability that the final selection is made by player B.

The answer in the back on the textbook was $$\frac{2}{5}$$ but I have no idea how they got that answer. Could someone please explain? Which formula are they using?

We can simply write down all the possibilities for the result:

A  B  A  B
y
r  y
r  r  y
r  r  r  y


We can then say that \begin{align}P(B \text{ wins})&=P(A\text{ picks red and then } B \text{ picks yellow}) \\ &\qquad + P(A \text{ picks red, }B\text{ picks red, }A\text{ picks red, }B\text{ picks yellow})\\ &=\frac35\times \frac24+\frac35\times\frac24\times\frac13\times\frac22\\ &=\frac6{20}+\frac{12}{120}\\ &=\frac3{10}+\frac1{10}\\ &=\frac4{10}\\\ &=\frac25\end{align}

There is no one magic 'formula' that'll immediately give you the answer. Instead, you'll have to think a bit about this problem, break it down, and go from there.

In particular, think of how $$B$$ could possibly select a yellow ball before $$A$$ does. Well, a moment's though reveals that there are two ways this could happen:

1. $$A$$ selects red, after which $$B$$ selects yellow

2. $$A$$ selects red, after which $$B$$ selects red, after which $$A$$ selects the last red, after which $$B$$ selects yellow.

Calculate the probabilities for each, and add them up.

I won't work this out further myself: you should be able to do this yourself!