If four dice are thrown simultaneously, then the probability that none of the dice shows the number 1 on its…
If four dice are thrown simultaneously, then the probability that none of the dice shows the number 1 on its face, is
$\frac{625}{1296}$
$\frac{125}{648}$
$\frac{1250}{1296}$
$\frac{625}{2592}$
Solution
Since there are 6 total samples. Hence the probability of getting number 1 on face $=\frac{1}{6}$
$\therefore$ Probability of not getting number 1 on face of a dice $=1-\frac{1}{6}=\frac{5}{6}$
Since 4 dice are thrown simultaneously. Hence probability of not getting number 1 on face of any of the dice
$
=\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)=\frac{625}{1296}
$