Q.3.5 - A First Course in Probability

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). $P\left(R_3\mid R_5\right)$

Step-by-Step Solution

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 1: 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.

Step 2: 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:

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

Step 3: Final Answer(Part a)

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

Step 4: 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

Step 5: 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:

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

Step 6: Final Answer (Part b)

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

Step 7: 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.

Step 8: 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):

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

Step 9: Final Answer (Part c)

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