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Question
A and B toss a coin alternately till one of them gets a head and wins the game. If A starts the game, find the probability that B will win the game.
Solution
\[P\left( B \text{ winning the game } \right) = P\left( {\text{head at }2}_{nd} \text{ turn } \right) + P\left( {\text{ head at } 4}_{th} \text{ turn } \right) + . . . \]
\[ = \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} + . . . \]
\[ = \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^4 + \left( \frac{1}{2} \right)^6 + \left( \frac{1}{2} \right)^8 + . . . \]
\[ = \frac{1}{4}\left[ 1 + \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^4 + \left( \frac{1}{2} \right)^6 + . . . \right]\]
\[ = \frac{1}{4}\left[ \frac{1}{1 - \frac{1}{4}} \right] \left[ \text{ For infinite } GP: 1 + a + a^2 + a^3 + . . . = \frac{1}{1 - a} \right]\]
\[ = \frac{1}{4} \times \frac{4}{3}\]
\[ = \frac{1}{3}\]
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