Q. 3.23 - A First Course in Probability

Question

A bag contains $\mathrm{a}$ white and $\mathrm{b}$ black balls. Balls are chosen from the bag according to the following method:

1. A ball is chosen at random and is discarded.

2. A second ball is then chosen. If its color is different from that of the preceding ball, it is replaced in the bag and the process is repeated from the beginning. If its color is the same, it is discarded and we start from step $2$.

In other words, balls are sampled and discarded until a change in color occurs, at which point the last ball is returned to the urn and the process starts anew. Let ${\mathrm{P}}_{\mathrm{a},\mathrm{b}}$ denote the probability that the last ball in the bag is white. Prove that

${\mathrm{P}}_{\mathrm{a},\mathrm{b}}=\frac{1}{2}$

Hint: Use induction on $\mathrm{K}=\mathrm{a}+\mathrm{b}$.

Step-by-Step Solution

Answer

The probability that the last ball in the bag is white, ${\mathrm{P}}_{\mathrm{a},\mathrm{b}}=\frac{1}{2}$

Step 1: Event

Beginning from $a$ white spheres plus $b$ black spheres, $E_{a,b}$ - the white sphere was that final one inside the container.

Every white spheres were selected initially $W$.

Every black spheres were selected initially $B$.

$a,b\in \mathrm{\mathbb{N}}=\{1,2,3,\dots \}$

Step 2: Show P a , b = 1 2

Base: to any or all $a,b\in \mathrm{\mathbb{N}}$ so $a+b=n=1$, which is that $a=b=1$ equation is correct, the balls inside urn was white by probability $\frac{1}{2}$

Assume that $n\in \mathrm{\mathbb{N}}$ and to any or all $a,b\in \mathrm{\mathbb{N}}$ so $a+b< n$.

                                                ${\mathrm{P}}_{\mathrm{a},\mathrm{b}}=\frac{1}{2}$

Consider $a,b\in \mathrm{\mathbb{N}}$ so $a+b=n:$

Because $W,B,W^c{B}^c$ were completely exclusive occurrences which which incorporates the whole outcome area, we may use the entire  probabilistic equation:

$P\left(E_{a,b}\right)=P\left(E_{a,b}\mid W^c{B}^c\right)P\left(W^c{B}^c\right)+P\left(E_{a,b}\mid B\right)P\left(B\right)+P\left(E_{a,b}\mid W\right)P\left(W\right)$

Step 3: show P E a , b &#8739; W c B c = P E i , j = 1 2 i + j &#60; n

If although neither a black nor white spheres were eliminated.  Especially at starting, will have at minimum a ball less, and method to get rid of the spheres begins.

The probability are easy when all spheres of a colour were taken:

                                    $P\left(E_{a,b}\mid W\right)=0P\left(E_{a,b}\mid B\right)=1$

                 $\begin{array}{c}\mathrm{P}\left(\mathrm{W}\right)=\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}}\times \frac{\mathrm{a}-1}{\mathrm{a}+\mathrm{b}-1}\cdot \cdots \frac{1}{\mathrm{b}+1}=\frac{\mathrm{a}!}{\frac{(\mathrm{a}+\mathrm{b})!}{\mathrm{b}!}}=\frac{\mathrm{a}!\mathrm{b}!}{(\mathrm{a}+\mathrm{b})!}\\ \mathrm{P}\left(\mathrm{B}\right)=\frac{\mathrm{b}}{\mathrm{a}+\mathrm{b}}\times \frac{\mathrm{b}-1}{\mathrm{a}+\mathrm{b}-1}\cdot \cdots \frac{1}{\mathrm{a}+1}=\frac{\mathrm{b}!}{\frac{(\mathrm{a}+\mathrm{b})!}{\mathrm{a}!}}=\frac{\mathrm{a}!\mathrm{b}!}{(\mathrm{a}+\mathrm{b})!}\\ \mathrm{P}\left(\mathrm{W}\right)=\mathrm{P}\left(\mathrm{B}\right)\end{array}$

So,

      $\mathrm{P}\left({\mathrm{E}}_{\mathrm{a},\mathrm{b}}\right)=\mathrm{P}\left({\mathrm{E}}_{\mathrm{a},\mathrm{b}}\mid {\mathrm{W}}^{\mathrm{c}}{\mathrm{B}}^{\mathrm{c}}\right)\mathrm{P}\left({\mathrm{W}}^{\mathrm{c}}{\mathrm{B}}^{\mathrm{c}}\right)+\mathrm{P}\left({\mathrm{E}}_{\mathrm{a},\mathrm{b}}\mid \mathrm{B}\right)\mathrm{P}\left(\mathrm{B}\right)+\mathrm{P}\left({\mathrm{E}}_{\mathrm{a},\mathrm{b}}\mid \mathrm{W}\right)\mathrm{P}\left(\mathrm{W}\right)$

                   $\begin{array}{r}=\frac{1}{2}\mathrm{P}\left({\mathrm{W}}^{\mathrm{c}}{\mathrm{B}}^{\mathrm{c}}\right)+0+1\times \mathrm{P}\left(\mathrm{W}\right)\\ =\frac{1}{2}\mathrm{P}\left({\mathrm{W}}^{\mathrm{c}}{\mathrm{B}}^{\mathrm{c}}\right)+\frac{1}{2}\times 2\mathrm{P}\left(\mathrm{W}\right)\end{array}$

                    $\begin{array}{r}=\frac{1}{2}\mathrm{P}\left({\mathrm{W}}^{\mathrm{c}}{\mathrm{B}}^{\mathrm{c}}\right)+\frac{1}{2}\times \left(\mathrm{P}\right(\mathrm{W})+\mathrm{P}(\mathrm{B}\left)\right)\\ \end{array}$

                     $=\frac{1}{2}\left(\mathrm{P}\left({\mathrm{W}}^{\mathrm{c}}{\mathrm{B}}^{\mathrm{c}}\right)+\mathrm{P}\left(\mathrm{W}\right)+\mathrm{P}\left(\mathrm{B}\right)\right)$

                     $=\frac{1}{2}\times 1$

                     $=\frac{1}{2}$