A jar contains 6 crayons, of which 3 are red, 2 are blue, and 1 is green. Jakob reaches into the jar and randomly removes 2 of the crayons. What is the probability that both of these crayons are red?
Solution 1

With 6 crayons in the box, there are $\frac{6 \times 5}{2}=15$ pairs of crayons that can be removed. (This is because there are 6 crayons that can be chosen first and 5 crayons that can chosen second; since there are 2 orders in which any specific pair of crayons can be chosen, we divide $6 \times 5$ by 2 .)

Using similar reasoning, with 3 red crayons in the box, there are $\frac{3 \times 2}{2}=3$ pairs of red crayons that can be removed.

Therefore, the probability that Jakob removes 2 red crayons is $\frac{3}{15}$ or $\frac{1}{5}$.

## Solution 2

We label the crayons in the box R1, R2, R3, B1, B2, G1.

The possible pairs that Jakob can choose are

$$
\text { R1R2, R1R3, R1B1, R1B2, R1G1, R2R3, R2B1, R2B2, R2G1, }
$$

$$
\text { R3B1, R3B2, R3G1, B1B2, B1G1, B2G1 }
$$

for a total of 15 pairs.

Of these 15 pairs, three pairs (R1R2, R1R3, R2R3) consist of two red crayons.

Therefore, the probability that Jakob removes 2 red crayons is $\frac{3}{15}$ or $\frac{1}{5}$.

## Solution 3

We remove the two crayons one after the other. In order for both crayons to be red, both the first and second crayons removed must be red.

When the first crayon is removed, the probability that it is red is $\frac{3}{6}$, since there are 6 crayons 3 , of which are red.

After 1 red crayon is removed, there are 5 crayons remaining, 2 of which are red.

Therefore, the probability that the second crayon removed is red is $\frac{2}{5}$.

Thus, the probability that both crayons are red is $\frac{3}{6} \cdot \frac{2}{5}=\frac{1}{5}$.