ONE MORE STEP TOWARD BETTER TOMORROW : chapter-9- Worked Out Examples

Thursday, 7 August 2014

chapter-9- Worked Out Examples – 2

Example: 3

What is the minimum number of times that a fair coin must be tossed so that the chances of getting at least one Head are greater than $99\%$ ?

Solution: 3

In a sequence of $n$ tosses , the probability of obtaining a Tail on every toss is

	$P$ (all tails in $n$ tossses) ${ =\underbrace{ \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \times \;\ldots \times \dfrac{1}{2}}_{\rm{Comment:1}}}$
	Comment:1 $n$ times
	$= \dfrac{1}{{{2^n}}}$

because at every toss, the probability of getting a Tail is $\dfrac{1}{2}$ , and also, all tosses are independent of each other.

Thus,

	$P({\rm{at\, least\, one\, head \,in}}\, n \, {\rm{tosses}}) \, =P\,=1-P\, (P{\rm{all\, tails\, in}}\, n \,{\rm{tosses}}) \,$
	$= 1 - \dfrac{1}{{{2^n}}}$

We want $P$ to be greater than $99 \%$ , or $0.99$ . Thus,

	$1 - \dfrac{1}{{{2^n}}} > .99$
	$\Rightarrow \,\,\,\, {2^n} > 100$
	$\Rightarrow \,\,\,\, n > 6$

Thus, a minimum of $7$ tosses are required.

Example: 4

Two persons $X$ and $Y$ are playing a game: They throw a coin alternately until one of them gets a Head and wins. How advantageous is it in such a game to make the first throw?

Solution: 4

Suppose that $X$ makes the first throw. Let us calculate the probability of $X$ winning the game.

Let ${H_x}$ , ${T_x}$ denote a Head and a Tail respectively obtained by $X$ . A similar notation follows for $Y$ . Now, $X$ will win the game in the following (mutually exclusive) sequences of tosses:

	SEQUENCE	$P$ (SEQUENCE)
	${H_x}$	$\dfrac{1}{2}$
	${T_x}{T_y}{H_x}$	$\dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{{{2^3}}}$
	${T_x}{T_y}{T_x}{T_y}{H_x}$	$\dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{{{2^5}}}$
	${T_x}{T_y}{T_x}{T_y}{T_x}{T_y}{H_x}$	$\dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{{{2^7}}}$
	$\vdots$	$\vdots$

Thus, the probability of $X$ winning the game is

	$P(X\,{\rm{wins}}) = \dfrac{1}{2} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^5}}} + \dfrac{1}{{{2^7}}} + \ldots \infty \;({\rm{a}}\;{\rm{G}}{\rm{.P}}.)$
	$= \dfrac{2}{3}$

This means that one who makes the first throw has twice the chance $\left( {\dfrac{2}{3}} \right)$ of winning than the other $\left( {\dfrac{1}{3}} \right)$

Thursday, 7 August 2014

chapter-9- Worked Out Examples – 2

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