A and B toss a coin alternatively, first to show a head being winner. If A starts game, chance of hi

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14.Probability

hard

$A$ 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

A

$5/8$

B

$1/2$

C

$1/3$

D

$2/3$

Solution

(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, …

$\therefore $ Probability of $A’s$ winning

$ = \frac{1}{2} + {\left( {\frac{1}{2}} \right)^2}.\frac{1}{2} + {\left( {\frac{1}{2}} \right)^4}.\frac{1}{2} + ……..$

$ = \frac{1}{2} + {\left( {\frac{1}{2}} \right)^3} + {\left( {\frac{1}{2}} \right)^5} + ……… = \frac{2}{3}$.

Standard 11

Mathematics

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