Urn without replacement i-th draw

probabilityprobability theory

Place blue, white and red balls in a urn. Prove that the probability that the ith ball taken from the box without replacement is:

$$ P(X_i=r) = \frac{r}{r+b+w}$$

There is a way to prove this result mathematically? Its true for $P(X_2 =r)$ (using total probability law), but how can I prove and interpret this result?

Thanks!

Best Answer

You may approach this in more than one way.

Let $n = r+b+w$.

"Direct way":

Each of the $n$ balls are equally likely to appear at the $i$-th position. There are $r$ red balls. So, you get $$P(X_i = r) = \frac{r}{n}= \frac{r}{r+b+w}$$

"Counting way:"

A possible way of counting is as follows:

  • There are $n!$ possible sequences to draw the $n$ balls without replacement.
  • There are $\color{red}{r}$ options to fix $\color{red}{\mbox{one}}$ red ball for the $i$-th position.
  • The remaining $n\color{red}{-1}$ balls can appear in $(n\color{red}{-1})!$ ways in the remaining positions.

All together: $$P(X_i = r)=\frac{\color{red}{r} \cdot (n\color{red}{-1})!}{n!}= \frac{\color{red}{r}}{n} = \frac{r}{r+b+w}$$

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