ONE MORE STEP TOWARD BETTER TOMORROW : chapter-23 Interesting Probability Questions

Thursday, 7 August 2014

chapter-23 Interesting Probability Questions

The subject of probability gives rise to some very interesting questions, which we believe everyone should be exposed to, even if to a minor extent. In this section, we’ll do precisely that. The examples will clearly show you the depth of this subject.

Example: 23

A drunk mathematician called $Pi$ is performing a random walk along the number line as follows. His original position is $x = 0$ . He takes a step forward or a step backward with equal probability. Thus, for example, if he is at $x = 1$ , he can go to $x = 2$ or $x = 0$ with equal probability.

His home is at $x = 3$ , while there is a pit at $\,x = - 2$ . What is the probability that he’ll reach home before falling into the pit?

Solution: 18

Assume the required probability to be $p$ . Thus, $p$ denotes the probability $P(E)$ where

$E$ : $Pi$ reaches home before the pit

Now, let us draw the probability tree that will make $E$ happen:

What should the events ${A_1}$ and ${A_2}$ be and what should be their probabilities? Once $Pi$ has reached $x = 1$ , the situation is as follows:

${A_1}$ should therefore be the event that $Pi$ reaches $x = 3$ before $x = - 2$ . Now comes the crucial insight. Compare Fig- $20$ with Fig- $18$ . The situations have been symmetrically reversed. In Fig - $18$ , $p$ was the probability that $Pi$ reaches $x = 3$ before $x = - 2$ , that is, $Pi$ travels $3$ units in one direction before travelling $2$ in the other. In Fig- $20$ , $P({A_1})$ is the probability that $Pi$ travels $2$ units in one direction before travelling $3$ in the other, so we simply have

$P({A_1}) = 1 - p$

Make sure you understand this argument carefully.

Coming to ${A_2}$ , $Pi$ has reached $x = - 1$ , he must return to $x = 0$ , because he must not reach $x = - 2$ . Once he is at $x = 0$ , he comes back to the starting configuration, which means that now his probability of reaching $x = 3$ before $x = - 2$ is again $p$ .

So let us complete the probability tree of Fig- $19$ now:

Thus (pay attention to how we write this!),

$p\,\, = \,\,\dfrac{1}{2}\,\, \times \,\,\mathop {(1 - p)}\limits_{\scriptstyle{\rm{Reaching}}\;\;x = 3\;\hfill\atop \scriptstyle{\rm{ from}}\;x = 1\hfill} \,\, + \,\,\dfrac{1}{2}\,\,\, \times \,\,\,\dfrac{1}{2}\,\, \times \mathop p\limits_{\scriptstyle{\rm{Back}}\;{\rm{to}}\hfill\atop \scriptstyle\,\,\,\,x = 0\hfill}$

$\ldots(1)$

This relation is a form of recursion. In the definition (or expression) of $p$ , we are using $p$ again. Thus, recursion is, in plain words, the process a ‘procedure’ goes through when one of the steps of the ‘procedure’ involves rerunning the procedure. In this example, the ‘procedure’ of calculating $p$ involved the step $(x = 0) \to (x = - 1) \to (x = 0),$ at which point we have to rerun the ‘procedure’ of calculating $p$ .

From $(1)$

	$p = \dfrac{1}{2} - \dfrac{1}{2}p + \dfrac{1}{4}p$
	$\Rightarrow\,\,\,\, \dfrac{5}{4}p = \dfrac{1}{2}$
	$\Rightarrow\,\,\,\, p = \dfrac{2}{5}$

Thus, there is a $40\%$ chance that $Pi$ will return home safely!

Recursions can also take the following form: suppose we have to calculate ${T_n}$ , the ${n^{th}}$ term of some sequence, and suppose we are able to express ${T_n}$ as some function of ${T_{n - 1}}$ , the previous term: