Expected value of the number of green balls chosen

Let $$X$$ be the random variable representing the number of green balls chosen, after choosing two balls uniformly at random without replacement from a bag containing exactly three green balls and two red balls. Find $$E[X]$$.

This was my thought process:

• $$X$$ can only be $$0$$, $$1$$ or $$2$$
• Find out the probability of X being each, and then calculate $$0\cdot\Pr(X = 0) + 1\cdot\Pr(X = 1) + 2\cdot\Pr(X = 2)$$, which equals $$E[X]$$

Probability $$X = 0$$:

$$(2/5) \cdot (1/4) = (1/10)$$

Probability $$X = 1$$:

$$(3/5) \cdot (2/4) = (3/10)$$

or

$$(2/5) \cdot (3/4) = (3/10)$$

Sum these two possible outcomes to get that $$\Pr(X = 1) = 3/5$$

Probability $$X = 2$$:

$$(3/5) \cdot (2/4) = (3/10)$$

E[X]

$$E[X] = 0\cdot(1/10) + 1\cdot(3/5) + 2\cdot(3/10) = 1.2$$

Is this process correct? Is there an easier way to solve this problem?

$$E[X]=\frac{nK}{N}=\frac{2(3)}{5}=1.2$$