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A die is thrown four times. The probability of getting perfect square in at least one throw is
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The correct answer is:
\(\mathrm{P}\) (getting perfect square in at least one throw) \(=1-\mathrm{P}\) (not getting any perfect square in any throw)
And among 1 to 6 , there are 2 perfect squares-1 and 4 .
\(=1-\left(\frac{4}{6} \times \frac{4}{6} \times \frac{4}{6} \times \frac{4}{6}\right)\)
\(=1-\frac{16}{81}\)
\(=\frac{65}{81}\)
And among 1 to 6 , there are 2 perfect squares-1 and 4 .
\(=1-\left(\frac{4}{6} \times \frac{4}{6} \times \frac{4}{6} \times \frac{4}{6}\right)\)
\(=1-\frac{16}{81}\)
\(=\frac{65}{81}\)
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