Probability Exercise asked by Dr. khalid in the Session preparation for math exam 2015

Exercise:

Consider $n-Urn$ with first urn has 2 black and 1 white and the other urns has each one back and one white.

Suppose we select one ball randomly from the first urn and we put it in the second urn repeating the process for other urns until the end of experiment

Let ${{W}_{k}}$ be the event of selecting white balls from urn $U_k$ where $1\le k\le n$

a) Find $P\left( {{W}_{1}} \right)$ ,$P\left( {{W}_{2}}/{{W}_{1}} \right)$ & $P\left( {{W}_{2}}/\overline{{{W}_{1}}} \right)$

b) Let ${{P}_{k}}=P\left( {{W}_{k}} \right)$

Show that , ${{P}_{k+1}}=\frac{1}{3}\left( {{P}_{k}}+1 \right)$

c) Show that , ${{P}_{n}}\to \frac{1}{2}$ as $n\to \infty $

Solution: First let’s illustrate the problem in drawing as follows:

a) $P\left( {{W}_{1}} \right)=\frac{1}{3}$

$P\left( {{W}_{2}}/{{W}_{1}} \right)=\frac{2}{3}$

$P\left( {{W}_{2}}/\overline{{{W}_{1}}} \right)=1-P\left( {{W}_{2}}/{{W}_{1}} \right)=1-\frac{2}{3}=\frac{1}{3}$

b) By induction the proof will be

For $k=1\Leftrightarrow {{P}_{2}}=\frac{1}{3}{{P}_{1}}+\frac{1}{3}=\frac{1}{3}P\left( {{W}_{1}} \right)+\frac{1}{3}=\frac{1}{3}\times \frac{1}{3}+\frac{1}{3}=\frac{1+3}{9}=\frac{4}{9}$

But ${{P}_{2}}=P\left( {{W}_{2}} \right)=P\left( {{W}_{2}}\cap {{W}_{1}} \right)+P\left( {{W}_{2}}\cap \overline{{{W}_{1}}} \right)$

Recall that $P\left( {{W}_{2}}/{{W}_{1}} \right)=\frac{P\left( {{W}_{2}}\cap {{W}_{1}} \right)}{P\left( {{W}_{1}} \right)}\Leftrightarrow P\left( {{W}_{2}}\cap {{W}_{1}} \right)=P\left( {{W}_{1}} \right)P\left( {{W}_{2}}/{{W}_{1}} \right)$

So ${{P}_{2}}=P\left( {{W}_{2}} \right)=P\left( {{W}_{1}} \right)P\left( {{W}_{2}}/{{W}_{1}} \right)+P\left( \overline{{{W}_{1}}} \right)P\left( {{W}_{2}}/\overline{{{W}_{1}}} \right)$

$\Rightarrow {{P}_{2}}=\frac{1}{3}\times \frac{2}{3}+\left( 1-\frac{1}{3} \right)\left( 1-\frac{2}{3} \right)=\frac{2}{9}+\frac{2}{9}=\frac{4}{9}$

Hence ${{P}_{2}}=\frac{1}{3}\left( {{P}_{1}}+1 \right)$ is true for $k=1$

Now suppose the statement is true for $k$ , ${{P}_{K}}=P\left( {{W}_{k}} \right)$

${{P}_{k+1}}=P\left( {{W}_{k+1}} \right)=P\left( {{W}_{k+1}}\cap {{W}_{k}} \right)+P\left( {{W}_{k+1}}\cap \overline{{{W}_{k}}} \right)$

      \[=P\left( {{W}_{k}} \right)P\left( {{W}_{k+1}}/{{W}_{k}} \right)+P\left( \overline{{{W}_{k}}} \right)P\left( {{W}_{k+1}}/\overline{{{W}_{k}}} \right)\]

      $={{P}_{k}}\left( \frac{2}{3} \right)+\left( 1-{{P}_{k}} \right)\left( \frac{1}{3} \right)={{P}_{k}}\left( \frac{2}{3} \right)+\frac{1}{3}-\frac{1}{3}{{P}_{k}}={{P}_{k}}\left( \frac{2}{3}-\frac{1}{3} \right)+\frac{1}{3}=\frac{1}{3}\left( {{P}_{k}}+1 \right)$

So the statement is also true for $k+1$ thus its true for any n

c) We have ${{P}_{k+1}}=\frac{1}{3}{{P}_{k}}+\frac{1}{3}$

Let $l=\underset{k\to \infty }{\mathop{\lim }}\,{{P}_{k}}=\underset{k\to \infty }{\mathop{\lim }}\,{{P}_{k+1}}$ also we have ${{v}_{n}}=\underset{k\to \infty }{\mathop{\lim }}\,{{P}_{k}}$ where ${{v}_{n}}={{P}_{n}}-l$

$\Leftrightarrow l=\frac{l}{3}+\frac{1}{3}\Leftrightarrow l\left( 1-\frac{1}{3} \right)=\frac{1}{3}\Leftrightarrow \frac{2}{3}l=\frac{1}{3}\Leftrightarrow l=\frac{1}{2}$

Hence ${{v}_{n}}={{P}_{n}}-\frac{1}{2}$

$r=\frac{{{v}_{n+1}}}{{{v}_{n}}}=\frac{{{P}_{n+1}}-\frac{1}{2}}{{{P}_{n}}-\frac{1}{2}}=\frac{\frac{1}{3}{{P}_{n}}+\frac{1}{3}-\frac{1}{2}}{{{P}_{n}}-\frac{1}{2}}=\frac{1}{3}$

So as result we obtain the following:

${{v}_{1}}={{P}_{1}}-\frac{1}{2}=\frac{1}{3}-\frac{1}{2}=\frac{-1}{6}$

${{v}_{2}}={{P}_{2}}-\frac{1}{2}=\frac{1}{3}{{P}_{1}}+\frac{1}{3}-\frac{1}{2}=\frac{1}{9}+\frac{1}{3}-\frac{1}{2}=\frac{-1}{18}$

Remark that if $\left( {{u}_{n}} \right)$ is geometric sequence such that ${{u}_{n}}=r{{u}_{n-1}}$

Where $r$ is the common ratio then${{u}_{n}}={{r}^{n-1}}{{u}_{1}}$

so ${{v}_{n}}={{r}^{n-1}}{{v}_{1}}={{\left( \frac{1}{3} \right)}^{n-1}}\left( \frac{-1}{6} \right)$ hence ${{P}_{n}}={{v}_{n}}+\frac{1}{2}={{\left( \frac{1}{3} \right)}^{n-1}}\left( \frac{-1}{6} \right)+\frac{1}{2}$

now as $n\to \infty \,\,,\,\,{{p}_{n}}\to \frac{1}{2}$






*
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Remark: the above sequence is neither arithmetic nor Geometric

Which is in the form ${{u}_{n}}=a{{u}_{n-1}}+b$ (*)

So we want to create from it a Geometric sequence as follows:

Take $\underset{n\to \infty }{\mathop{\lim }}\,{{u}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,{{u}_{n-1}}=l$

so (*) will be $l=al+b\Leftrightarrow l=\frac{b}{1-a}$

Thus ${{v}_{n}}={{u}_{n}}-l$  with the ratio $r=\frac{{{v}_{n+1}}}{{{v}_{n}}}$ i.e ${{v}_{n}}={{r}^{n-1}}{{v}_{1}}$