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A bag contains 9 red marbles, 5 blue marbles and 7 green marbles. Need to select one marble at a time without replacement. What is the probability that the first marble is blue, the second is blue or green an the third is red.

My Solution:

$= P(\text{blue}) * P(\text{blue} \cup \text{green}) * P(\text{red})$

$\displaystyle = {5\over 21} * ({4\over 20} + {7\over 20}) * ({9\over 19})$

Does that sound that right?

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    $\begingroup$ It's fine. If the second choice was blue or red, one would have to change the analysis. $\endgroup$ May 25, 2014 at 2:40

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That looks like the right answer. These are conditional probabilities, so

\begin{align*} & \mathrm{Pr}(\text{marble 1 is blue} \cap \text{marble 2 is blue or green} \cap \text{marble 3 is red}) \\ &= \mathrm{Pr}(\text{marble 1 is blue}) \times \mathrm{Pr}(\text{marble 2 is blue or green} \mid \text{marble 1 is blue}) \times \mathrm{Pr}(\text{marble 3 is red} \mid \text{marble 1 is blue} \cap \text{marble 2 is blue or green}) \\ &= \frac{5}{9+5+7} \times \frac{(5+7)-1}{(9+5+7)-1} \times \frac{9}{(9+5+7)-2}\\ &= \frac{33}{532}. \end{align*}


Alternatively, we can use a counting argument. There are $3! \binom{9+5+7}{3}=7980$ ordered $3$-tuples of $\{r_1,\ldots,r_9,b_1,\ldots,b_5,g_1,\ldots,g_7\}$. Of these, precisely $5 \times (5+7-1) \times 9=495$ satisfies the blue-blue/green-red pattern. We have $$\frac{5 \times (5+7-1) \times 9}{3! \binom{9+5+7}{3}}=\frac{33}{532}.$$

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