# Drawing 2 marbles from a box with 3 red, 3 purple, 5 green, and 7 blue marbles.

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?

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.

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.

• 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. – Marconius Nov 17 '15 at 21:35