# Selecting multiple balls of same color from different urns

Urn I contains 10 balls: 4 red and 6 blue. Urn II contains 20 balls: 16 red and 4 blue. A single ball is drawn from urn I, and two balls are drawn from urn II. What is the probability that all three balls are the same color?

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What have you tried? –  Alex Becker Jan 11 '12 at 20:13

Numer of all possible ways of selecting 1 ball from urn 1 and 2 balls from urn 2 is ${\left( \begin{gathered} 10 \\ 1 \\ \end{gathered} \right)\left( \begin{gathered} 20 \\ 2 \\ \end{gathered} \right)}$.

Number of ways to select 1 red ball from urn 1 and 2 red balls from urn 2 is ${\left( \begin{gathered} 4 \\ 1 \\ \end{gathered} \right)\left( \begin{gathered} 16 \\ 2 \\ \end{gathered} \right)}$.

Number of ways to select 1 blue ball from urn 1 and 2 blue balls from urn 2 is ${\left( \begin{gathered} 6 \\ 1 \\ \end{gathered} \right)\left( \begin{gathered} 4 \\ 2 \\ \end{gathered} \right)}$.

Hence, number of ways of selecting 1 ball from urn 1 and 2 balls from urn 2 such that they are of the same color is ${\left( \begin{gathered} 4 \\ 1 \\ \end{gathered} \right)\left( \begin{gathered} 16 \\ 2 \\ \end{gathered} \right) + \left( \begin{gathered} 6 \\ 1 \\ \end{gathered} \right)\left( \begin{gathered} 4 \\ 2 \\ \end{gathered} \right)}$ since they are mutually exclusive.

Since probability of selecting balls of the same color is number of ways that can be done divided by number of all possible ways of selecting 1 ball from urn 1 and 2 balls from urn 2, we have that wanted probability is

$\frac{{\left( \begin{gathered} 4 \\ 1 \\ \end{gathered} \right)\left( \begin{gathered} 16 \\ 2 \\ \end{gathered} \right) + \left( \begin{gathered} 6 \\ 1 \\ \end{gathered} \right)\left( \begin{gathered} 4 \\ 2 \\ \end{gathered} \right)}} {{\left( \begin{gathered} 10 \\ 1 \\ \end{gathered} \right)\left( \begin{gathered} 20 \\ 2 \\ \end{gathered} \right)}} = \frac{{129}} {{475}}$

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In this case, we can compute directly. Imagine drawing from Urn 1, then sequentially from Urn 2. Using the obvious shorthand, we want the probability of RRR plus the probability of BBB. This is $$\frac{4}{10}\cdot\frac{16}{20}\cdot\frac{15}{19}\:+\: \frac{6}{10}\cdot\frac{4}{20}\cdot\frac{3}{19}.$$

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