A die is thrown four times. The probability of getting perfect square in at least one throw is
A die is thrown four times. The probability of getting perfect square in at least one throw is
$\frac{58}{61}$
$\frac{16}{81}$
$\frac{65}{81}$
$\frac{23}{81}$
Solution
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 $=1-\frac{16}{81}=\frac{65}{81}$