# 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?

Nov 27, 2017

$\frac{3}{11}$

#### 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 $\frac{5}{12}$. After we draw the red, there are 4 green out of 11 sweets (since the red is out), and so the probability of drawing the green is $\frac{4}{11}$. After we draw the green, there are 3 yellow out of 10 sweets, leaving the probability of drawing the yellow as $\frac{3}{10}$.

Using the Multiplication Principle, we can find the probability of the sequence RGY to be: $\frac{5}{12} \cdot \frac{4}{11} \cdot \frac{3}{10} = \frac{60}{1320} = \frac{1}{22}$.

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 $\frac{5}{12} \cdot \frac{3}{11} \cdot \frac{4}{10} = \frac{1}{22}$.

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: $\frac{1}{22}$. Therefore, the total overall probability is $6 \cdot \frac{1}{22} = \frac{3}{11}$

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 ""_12C_3 ways to choose 3 sweets without replacement from the 12 sweets. However, we're interested in a combination which results in one of each.

Considering that there are 5 red sweets, there are ""_5C_1 ways to choose a red sweet. Likewise there are ""_4C_1 ways to choose a green sweet and ""_3C_1 ways to choose a yellow sweet. Thus, our probability of drawing one of each color is:

$\left({\text{_5C_1 * ""_4C_1 * ""_3C_1)/(}}_{12} {C}_{3}\right) = \frac{5 \cdot 4 \cdot 3}{220} = \frac{60}{220} = \frac{6}{22} = \frac{3}{11}$