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.
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 Xwinning 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:
{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}}}
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)

No comments: