Four fair dice $D_{1}, D_{2}, D_{3}$ and $D_{4}$ each having six faces numbered $1, 2, 3, 4, 5$ and $6$ are rolled simultaneously. The probability $D_{4}$ shows a number appearing on one of $D_{1}, D_{2}$ and $D_{3}$ is

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$\frac{127}{216}$

$\frac{108}{216}$

$\frac{125}{216}$

$\frac{91}{216}$

answer is A.

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Detailed Solution

$P (D_{4}$ shows number on one of $D_{1}, D_{2}$ and $D_{3})$

$= 1 - P (D_{4}$ shows a number different from $D_{1}, D_{2}$ and $D_{3}$ $)$

$= 1 - \frac{^{6} C_{1} (5^{3})}{6^{4}} = 1 - \frac{125}{216} = \frac{91}{216}$

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