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
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!


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