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A person tossing a biased coin indefinitely wins the game by getting head for the first time. The probability that he wins the game in odd number of tosses is $3 / 4$. If 5 such coins are tossed at a time then the probability that head appears on all the coins is
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The correct answer is:
$\frac{32}{243}$
Let the probability getting head $=p$
probability of not getting head $=1-p$
Given, he wins the game in odd number of tosses
$\begin{gathered}
\therefore \quad p+(1-p)^2 p+(1-p)^4 p+\ldots=\frac{3}{4} \\
\frac{p}{1-(1-p)^2}=\frac{3}{4} \\
4 p=3-3\left(1-2 p+p^2\right)=6 p-3 p^2 \\
3 p^2=2 p \Rightarrow p=\frac{2}{3}
\end{gathered}$
5 coins are tossed at a time the probability of head appears is
${ }^5 C_5\left(\frac{2}{3}\right)^5=\frac{32}{343}$
probability of not getting head $=1-p$
Given, he wins the game in odd number of tosses
$\begin{gathered}
\therefore \quad p+(1-p)^2 p+(1-p)^4 p+\ldots=\frac{3}{4} \\
\frac{p}{1-(1-p)^2}=\frac{3}{4} \\
4 p=3-3\left(1-2 p+p^2\right)=6 p-3 p^2 \\
3 p^2=2 p \Rightarrow p=\frac{2}{3}
\end{gathered}$
5 coins are tossed at a time the probability of head appears is
${ }^5 C_5\left(\frac{2}{3}\right)^5=\frac{32}{343}$
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