# Probability of picking two marbles each from two colors when selecting $4$ marbles out of $30$ marbles

I have $$\ 30$$ marbles. $$\ 25$$ are white, $$\ 3$$ are blue and $$\ 2$$ are red. same color marbles are identical.

If I pick randomly and without replacement $$\ 4$$ marbles, what is the probability that I'll pick two each two of two colors?

Trying to make it easier, I assumed all marbles are different, so there are $$\ 30 \cdot 29 \cdot 28 \cdot 27$$ ways to pick them and then number of options for :

Picking $$2$$ blue and $$2$$ red marbles are $$\ {3 \choose 2}{25 \choose 2} \cdot 4!$$ options.

Picking $$2$$ blue and $$2$$ white marbles are $$\ {3 \choose 2}{2 \choose 2} \cdot 4!$$ options.

Picking $$2$$ white and $$2$$ red marbles are $$\ {25 \choose 2}{2 \choose 2 }\cdot 4!$$.

The three events are mutually exclusive, so I should be able to just add them all together but that's the wrong answer. Any suggestions?

• I assume 2 white and 2 white isn't an option? – Arthur Oct 31 '18 at 12:39
• Do you mean two each of two colors? – N. F. Taussig Oct 31 '18 at 12:40
• Correct. Sorry I try to be clear as I could, I translated this question to english. 4 marbles, 2 different colors, 2 marlbes of each color – bm1125 Oct 31 '18 at 12:40
• So what is the final answer you get, and what is the answer you should've gotten? – Arthur Oct 31 '18 at 13:13
• Calculating again, I suddenly got different answer on my calculator which fit one of the possible answers ( $\ \frac{401}{9135}$ ). But I'm not sure though if it is the answer? – bm1125 Oct 31 '18 at 13:19

Correct answer is $$\frac{401}{9135}$$. It is calculated as follows:$$\frac{\binom{25}{2}*\binom{3}{2}}{\binom{30}{4}}+\frac{\binom{25}{2}*\binom{2}{2}}{\binom{30}{4}}+\frac{\binom{3}{2}*\binom{2}{2}}{\binom{30}{4}}$$
You are drawing a random $$4$$-element subset from a set $$S$$ containing $$30$$ elements. (Even though some elements of $$S$$ look alike the elements of $$S$$ are "secretly" numbered: $$1$$$$25$$ for the white ones, $$26$$$$28$$ for the blue ones, and $$29$$$$30$$ for the red ones.) There are three kinds of "good" subsets. You have counted them, but for no reason multiplied the numbers by $$4!$$. (Note that the order in which the four marbles are drawn plays no rôle. All four are put in a smaller bag.) The probablity you are after is the total number of "good" subsets divided by the total number of all $$4$$-element subsets.
• I multiplied by $\ 4!$ because as you said I have the elements of S are "secretly" numbered and therefore I want to count both $\ R,R,B,B$ and $\ B,B,R,R$ . I mean maybe if I had chosen instead of $\ 30 \cdot 29 \cdot 28 \cdot 27$ a $\ 30!/(26!4!)$ I wouldn't have to multiply by $\ 4!$ any event? – bm1125 Oct 31 '18 at 13:37