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?
Yes, it is correct.
This is known as hypergeometric distribution,
$$E[X]=\frac{nK}{N}=\frac{2(3)}{5}=1.2$$
The formula can be understood as linearity of expectation.