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There are 10 coins in a box out of which 8 are normal and the remaining are with heads on both sides. A coin is chosen at random from the box and tossed 6 times. If it shows heads each time, then the probability that the selected coin has head on both sides is
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The correct answer is:
$\frac{16}{17}$
$P($ both side head $)=\frac{2}{10}$
$P($ Fair coin $)=\frac{8}{10}$
$\begin{aligned} & P\left(\frac{\text { Head }}{\text { Fair }}\right)=\left(\frac{1}{2}\right)^6 \\ & P\left(\frac{\text { Head }}{\text { Unfair }}\right)=1\end{aligned}$
Probability $=\frac{\frac{2}{10} \times 1}{\frac{2}{10} \times 1+\left(\frac{1}{2}\right)^6 \times \frac{8}{10}}=\frac{16}{17}$
$P($ Fair coin $)=\frac{8}{10}$
$\begin{aligned} & P\left(\frac{\text { Head }}{\text { Fair }}\right)=\left(\frac{1}{2}\right)^6 \\ & P\left(\frac{\text { Head }}{\text { Unfair }}\right)=1\end{aligned}$
Probability $=\frac{\frac{2}{10} \times 1}{\frac{2}{10} \times 1+\left(\frac{1}{2}\right)^6 \times \frac{8}{10}}=\frac{16}{17}$
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