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In a simultaneous throw of three coins, what is the probability of getting at least 2 tails
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The correct answer is:
$\frac{1}{2}$
Total ways are 8 and fayourable ways are 4
$S=\{H H H, H H T \ldots \ldots . T T T\}$.
Hence probability $=\frac{4}{8}=\frac{1}{2}$.
$S=\{H H H, H H T \ldots \ldots . T T T\}$.
Hence probability $=\frac{4}{8}=\frac{1}{2}$.
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