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I'm going through some of my probability notes and came across this problem which is rather simple:

A box contains 9 white balls and 16 blue balls. Draw three without replacement. Find the expected number of white balls.

We proceed as follows: $$E[White balls] = n*p$$ where $n = 3$ and $p = \frac{white\ balls}{total\ balsls} = \frac{9}{25}$ We can now apply the expectation of a binomial random variable which we set up above $$E[White balls] = 3 * \frac{9}{25} = 1.08$$

However, looking at this now I notice that a binomial random variable is a sum of independent Bernoulli Trials, which isn't the case here. Every time we draw without replacement we affect the probabilities of the next draw. So we can't possibly use the expectation of a binomial random variable here. Can we?

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  • $\begingroup$ Linearity of expectation? $\endgroup$ – Brian Tung Jun 22 '20 at 2:40
  • $\begingroup$ To answer your final question: You can, but it's sort of an accident (via the linearity of expectation). The usual distribution in question is hypergeometric, but since you only need the expected number... $\endgroup$ – Brian Tung Jun 22 '20 at 2:41
  • $\begingroup$ Wikipedia plot summary for hypergeometric distribution. Overview of linearity of expectation. $\endgroup$ – Brian Tung Jun 22 '20 at 2:51
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As mentioned above by Brian Tung, the distribution of the number of white balls is the hypergeometric distribution, but the expectation of the distribution is the same as the corresponding binomial distribution.

To show this using linearity of expectation (again as mentioned above), let $X_j$ equal 1 if the $j$th ball is white and 0 otherwise. We may apply a symmetry argument to conclude that $$E[X_j] = p$$ for any $j$. Thus, the desired expectation is given by $$E[X_1 + X_2 + X_3] = E[X_1] + E[X_2] + E[X_3] = 3p.$$

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