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If four dice are thrown simultaneously, then the probability that none of the dice shows the number 1 on its face, is
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Verified Answer
The correct answer is:
$\frac{625}{1296}$
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
$$
=\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)=\frac{625}{1296}
$$
$\therefore$ 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
$$
=\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)=\frac{625}{1296}
$$
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