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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?

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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.

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