# Probability: basic question and concept

I have always been struggling with the problem, in particular, I usually have great difficulty in differenting when should I multiply n! to take care of the ordering, and when should I not do so. For simple question, this might not be an issue; but if this basic concept is not understood throughly, this will extend to more complicate question. Because of this, I hope you guys can correct my understanding of basic probably.

For instance, a very basic question:

12 balls in a pool, 7 green, 2 yellow, 3 orange. What is the probability that of drawing 2 yellow and 2 orange?

My attempt:

First Approach:

the simplest way is to use view each draw is independent (this approach I have no problem): $$\frac{3}{12}\cdot\frac{2}{11}\cdot\frac{2}{10}\cdot\frac{1}{9}\cdot4!$$ 4! because such "independent draw" could happen 4! times. So this approach I have no problem.

Second Approach: \begin{align*} P(\text{2 yellow AND 2 orange})&=\frac{\text{# of ways you can draw 2 yellow AND 2 orange}}{\text{# of ways draw 4 balls}} \\ &=\frac{\binom{3}{2}\binom{2}{2}}{\binom{12}{4}} \end{align*}

my understanding:

• $\binom{3}{2}$ is the number of ways you can form 2 orange out of the 3, where the ordering is not important.
• you mutiply by $\binom{2}{2}$ is because for each of $\binom{3}{2}$ ways, you can have $\binom{2}{2}$ ways to combine them with the yellow balls.

It terms out that I am missing a factor of 4. It turns out that I am having true difficulty with this type of problem. What can I do to improve?

If you count the number of ways of drawing the balls such that balls of the same color are distinguishable and such that order does not matter, then you should have $$\frac{\binom{3}{2} \cdot 1}{\binom{12}{4}},$$ as you have written.
The first approach is incorrect. $\frac{3}{12} \cdot \frac{2}{11} \cdot \frac{2}{10} \cdot \frac{1}{9}$ is the probability of choosing orange first, orange second, yellow third, and yellow fourth, as well as any other ordering. You only want to multiply this by $\frac{4}{2!2!}$ (the number of ways of arranging "orange, orange, yellow, yellow" in a row) instead of $4!$ (number of permutations of 4 distinguishable objects). The erroneous factor of $4$ comes because each arrangement of "orange, orange, yellow, yellow" is counted 4 times instead of once because you distinguished between the two oranges and the two yellows when multiplying by $4!$.
• @kuku As I wrote above, the approach with the $\binom{3}{2}$ involves the balls of the same color being distinguishable. This is ok as long as you assume the same thing when computing the denominator. – angryavian Sep 14 '14 at 11:15