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A box contains 3 red, 3 purple, 5 green, and 7 blue marbles. 2 marbles are selected from the box without replacement. What is the probability that you choose both marbles to be red or both marbles to be purple.

So far, I have figured out (I think):

$$\text{ Probability of both red }= \frac{3}{\binom{18}{2}} = \frac{3}{(\frac{18!}{2!16!})} = \frac{3}{153}$$

Is this correct?

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You are correct. Another way to compute it:

Let $A_i$ be the event in which the $i$-th ball you grab is red. For the first ball you choose, you have $3$ red marbles out of a total of $18$. Therefore,

$$P(A_1) = \frac{3}{18}.$$

For the second marble you grab, given that you already took a red one, that is, given $A_1$, you have $2$ red balls out of a total of $17$, then

$$P(A_2 \mid A_1) = \frac{2}{17}.$$

Finally,

$$P(A_1 \cap A_2) = P(A_1)P(A_2 \mid A_1) = \frac{3}{18}\frac{2}{17} = \frac{1}{51}.$$

You can use the same reasoning to compute for the purple ones and then add those probabilities.

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Actually, the total number of ways for both marbles to be red or both to be purple is $3*2+3*2=12$. Then it would be $\frac{12}{\dbinom{18}{2}}$=$\frac{12}{153}$=$\boxed{\frac{4}{51}}$, I believe. So for just both red, it would be 6 ways, not 3.

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    $\begingroup$ Division by $\binom{18}{2}$ implies that combinations, not permutations, are being considered. So there are $3$ combinations of red balls, not $3\times2$. Or if you consider perms, then $\frac{3\times2}{18\times17}$, which is $1/51$. In your interpretation of the question, you are correct to add "red" and "purple" probabilities as these are mutually exclusive. $\endgroup$ – Marconius Nov 17 '15 at 21:35

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