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 -18p was the probability that Pi reaches  x = 3  beforex =  - 2, that is, Pi travels 3 units in one direction before travelling 2 in the other. In Fig-20P({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:
{T_n} = f({T_{n - 1}})
This is also a recursion relation, for now we can use this relation repeatedly to descend, that is, to reach some {T_m} that is calculable:
{T_n} = f({T_{n - 1}}) = f(f({T_{n - 2}})) = f(f(f({T_{n - 3}}))) = \ldots  = f(f(f(\ldots {T_m}))))\ldots )
For example, {T_m} could be the first term of the sequence. Thus, here we are rerunning the ‘procedure’ f.

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