3
$\begingroup$

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?

$\endgroup$
8
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.