If four dice are thrown simultaneously, then the probability that none of the dice shows the number 1 on its face, is

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 $$ =\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)=\frac{625}{1296} $$