# A bag contains 12 sweets, of which 5 are red, 4 are green and 3 are yellow. 3 sweets are chosen without replacement. What is the probability that he chooses 1 sweet of each color?

##### 1 Answer

#### Explanation:

There are 2 ways to look at this problem: As a "standard" probability question, or as a combinations question.

**"Standard" Probability**

Consider the possibility that the draw takes place and we draw a red, followed by a green, followed by a yellow. Since there are originally 12 sweets, of which 5 are red. we know the probability of drawing a red is

Using the Multiplication Principle, we can find the probability of the sequence RGY to be:

However, that's not the only path towards our desired end. We could also draw the sweets in the order RYG and still be acceptable. In that case, following the same logic, the probability of going RYG is

Even still, there are other orders we could draw in to work out an acceptable result. In fact, there are 6 total: RGY, RYG, GRY, GYR, YRG, and YCR. Because of the commutative properties of fraction multiplication, it turns out that all 6 of these patterns have the exact same individual probabilities:

**Combinations**

As an alternative, we can solve this as a combinations problem. Since the order of the drawing of the sweets does not matter - only that we get exactly 1 of each of red, yellow, and green.

Overall, regardless of what is drawn, there are

Considering that there are 5 red sweets, there are