A and B toss a coin alternatively, the first | vedclass

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Question

(d) The chance of head $ = \frac{1}{2}$ and not of head $ = \frac{1}{2}$

Since $A$ has first throw, he can win in the first, third, &hellip;

$

Vedclass · Accepted Answer

$2/3$

Vedclass · Answer

(d) The chance of head $ = \frac{1}{2}$ and not of head $ = \frac{1}{2}$

Since $A$ has first throw, he can win in the first, third, &hellip;

$	herefore $ Probability of $A&rsquo;s$ winning

$ = \frac{1}{2} + {\left( {\frac{1}{2}} ight)^2}.\frac{1}{2} + {\left( {\frac{1}{2}} ight)^4}.\frac{1}{2} + ........$

$ = \frac{1}{2} + {\left( {\frac{1}{2}} ight)^3} + {\left( {\frac{1}{2}} ight)^5} + ......... = \frac{2}{3}$.

$A$ and $B$ toss a coin alternatively, the first to show a head being the winner. If $A$ starts the game, the chance of his winning is

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