A die is thrown four times. The probability o | vedclass

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(C) The possible outcomes on a die are ${1, 2, 3, 4, 5, 6}$.The perfect squares among these are ${1, 4}$.Thus,the probability of getting a perfect squ

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$\frac{65}{81}$

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(C) The possible outcomes on a die are ${1, 2, 3, 4, 5, 6}$.The perfect squares among these are ${1, 4}$.Thus,the probability of getting a perfect square in a single throw is $p = \frac{2}{6} = \frac{1}{3}$.The probability of not getting a perfect square in a single throw is $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.For $n = 4$ independent throws,the probability of not getting a perfect square in any of the four throws is $q^4 = (\frac{2}{3})^4 = \frac{16}{81}$.The probability of getting a perfect square in at least one throw is $1 - P(	ext{no perfect square}) = 1 - \frac{16}{81} = \frac{65}{81}$.

$A$ die is thrown four times. The probability of getting a perfect square in at least one throw is

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