tgt

## 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 $\left( {\dfrac{1}{3}} \right)$