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  • In a game, two players X and Y are tossing a coin alternately. Whoever gets a 'head' first, wins the game and the game is terminated. Assuming the player X winning the game i

In a game, two players X and Y are tossing a coin alternately. Whoever gets a 'head' first, wins the game and the game is terminated. Assuming the player X winning the game is

Option: 1

\frac{1}{3}


Option: 2

\frac{1}{3}


Option: 3

\frac{2}{3}


Option: 4

\frac{3}{4}


Answers (1)

best_answer

Let

\mathrm{P=P(H)=\frac{1}{2}, q=P(T)=\frac{1}{2} \text {. }}

So A can win the game either in 1st or in 3rd or in 5th toss and so on ...

\begin{aligned} \therefore \quad \mathrm{P(A)} & =\mathrm{p+q^2 p+q^4 p+\ldots} \\ & = \mathrm{p\left[1+q^2+q^4+\ldots \infty\right]} \end{aligned}

                       \mathrm{=\frac{p}{1-q^2}=\frac{2}{3}}

Game 

Posted by

Pankaj

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