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If three dice are thrown together, then the probability of getting 5 on at least one of them is
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The correct answer is:
$\frac{91}{216}$
Required probability
$={ }^3 C_1\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^2$ $+{ }^3 C_2\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)+{ }^3 C_3\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^0=\frac{91}{216}$
$={ }^3 C_1\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^2$ $+{ }^3 C_2\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)+{ }^3 C_3\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^0=\frac{91}{216}$
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