91Ó°ÊÓ

$r$ red, $w$ white balls.

choose one by one in a random order.

$R_i-i$-th drawn ball is red.

Random order means that every of the $(r+w)$ ! permutations is equally likely to be the order of drawing the balls.

Also, that means that every of the $(r+w)$balls is equally likely to be the $i$-th drawn.

As there are $r$of them:

The required probability is $P\left(R_i\right)=\frac{r}{r+w}$.

$r$ red, $w$ white balls

choose one by one in a random order

$R_i-i$-th drawn ball is red

Reduction of sample space: If $R_3$ one of the red balls is the third drawn one. the remaining $r+w-1$ balls are randomly drawn. Each of them is equally likely to be the 5th. As $r-1$ of them are red:

The required probability is $P\left(R_5âˆ\pounds R_3\right)=\frac{r-1}{r+w-1}$.

$r$ red, $w$ white balls.

choose one by one in a random order.

$R_i-i$-th drawn ball is red.

If it is known that the fifth ball is red, the order no longer makes a difference, therefore c) is the same as b):

$P\left(R_3âˆ\pounds R_5\right)=\frac{r-1}{r+w-1}$.

_{The required probability is$P\left(R_3âˆ\pounds R_5\right)=\frac{r-1}{r+w-1}$.}