Question

An urn has r red and w white balls that are randomly removed one at a time. Let $Ri$be the event that the ith ball removed is red. Find

a). $P\left(R_i\right)$

b). $P\left(R_5\mid R_3\right)$

c).$P\left(R_3\mid R_5\right)$

Short answer

The required probabilities are,

a) $P\left(R_i\right)=\frac{r}{r+w}$

b) $P\left(R_5\mid R_3\right)=\frac{r-1}{r+w-1}$

c) $P\left(R_3\mid R_5\right)=\frac{r-1}{r+w-1}$

Step-by-step solution

Given Information (part a)

$r$ red, $w$ white balls.

choose one by one in a random order.

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

Explanation (Part a)

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:

$P\left(R_i\right)=\frac{r}{r+w}$

Final Answer(Part a)

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

Given Information (Part b)

$r$ red, $w$ white balls

choose one by one in a random order

$R_i-i$-th drawn ball is red

Explanation (Part b)

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:

role="math" localid="1647945836225" $P\left(R_5\mid R_3\right)=\frac{r-1}{r+w-1}$

Final Answer (Part b)

The required probability is $P\left(R_5\mid R_3\right)=\frac{r-1}{r+w-1}$.

Given Information (Part c)

$r$ red, $w$ white balls.

choose one by one in a random order.

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

Explanation (Part c)

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

$P\left(R_3\mid R_5\right)=\frac{r-1}{r+w-1}$.

Final Answer (Part c)

_{The required probability is$P\left(R_3\mid R_5\right)=\frac{r-1}{r+w-1}$.}