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(B) The total outcomes on a single die are $\{1, 2, 3, 4, 5, 6\}$.There are $2$ perfect squares in this set,which are $1$ and $4$.So,the probability o

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$\frac{65}{81}$

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(B) The total outcomes on a single die are $\{1, 2, 3, 4, 5, 6\}$.There are $2$ perfect squares in this set,which are $1$ and $4$.So,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 $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.Since the die is thrown $4$ times,the probability of not getting a perfect square in any of the $4$ 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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