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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:
$\frac{65}{81}$
From 1 to 6 , we have 1 and 4 as perfect squares.
Probability of getting perfect square is one throw of a die
$$
=\frac{2}{6}=\frac{1}{3}
$$
$\therefore$ Probability of not getting perfect square in 4 throws of a die
$$
=\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}=\frac{16}{81}
$$
$\therefore$ Required probability $=1-\frac{16}{81}=\frac{65}{81}$
Probability of getting perfect square is one throw of a die
$$
=\frac{2}{6}=\frac{1}{3}
$$
$\therefore$ Probability of not getting perfect square in 4 throws of a die
$$
=\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}=\frac{16}{81}
$$
$\therefore$ Required probability $=1-\frac{16}{81}=\frac{65}{81}$
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