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A box contain $A$ white and $B$ black balls and $C$ balls are drawn, then the expected value of the number of white balls drawn is ?

The answer is $\large \frac{ca}{a+b}$. How to approach this one?

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FYI: The number of white balls is said to have a hypergeometric distribution: en.wikipedia.org/wiki/Hypergeometric_distribution. –  Mike Spivey Jan 27 '11 at 17:09
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2 Answers

up vote 6 down vote accepted

These calculations are often made clearer by using indicator random variables. For $1\leq i\leq C$, define $Z_i$ by setting $Z_i=1$ if the $i$th ball drawn is white and $Z_i=0$ otherwise. Then the total number of white balls is $\sum_{i=1}^C Z_i$ and by linearity of expectation we have $$ E(\mbox{ white balls })=E\left(\sum_{i=1}^C Z_i\right)=\sum_{i=1}^C E(Z_i).$$

The expectation of an indicator random variable is just the probability of the event it indicates, so $E(Z_i)=P(i\mbox{th ball is white})=A/(A+B)$.

Therefore we find that $$ E(\mbox{ white balls })=\sum_{i=1}^C E(Z_i)={C A\over A+B}.$$

Notice that we did not need to calculate the probability of getting one, two, three, etc. white balls.

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For those interested, just to elucidate on the "non-smart" method of attacking this problem, the probability of drawing exactly $k$ white balls in $C$ draws is $$ \frac{ {A \choose k}{B \choose C-k}} { {A+B \choose C} } $$ and so the expected number of white balls is given by

$$\sum_{k=0}^C \frac{ k {A \choose k}{B \choose C-k}} { {A+B \choose C} } = \frac{AC}{A+B}.$$

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