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 that $D_{4}$ shows a number appearing on one of $D_{1}, D_{2}$ and $D_{3}$ is

\frac{91}{216}

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\frac{108}{216}

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\frac{125}{216}

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\frac{127}{216}

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Solution

The correct option is A $\frac{91}{216}$
Four fair dices rolled simultaneously.

Total number of outcomes $= 6 \times 6 \times 6 \times 6$

Suppose, the out come of the last one does not appear in the first three.

The number of such outcomes possible $= 6 \times 5^{3}$

$(∵ D_{1}$ can show any of 6 numbers and first three can show any of the remaining 5 numbers)

Hence, probability that $D_{4}$ does not show a number shown by any of $D_{1}, D_{2}$ or $D_{3}$

$= \frac{6 \times 5^{3}}{6^{4}}$

So, the required probability

$= 1 - \frac{6 \times 5^{3}}{6^{4}} = \frac{91}{216}$

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Four fair dice D1, D2, D3 and D4, each having six faces numbered 1,2,3,4,5 and 6, are rolled simultaneously. The probability that D4 shows a number appearing on one of D1, D2 and D3 is

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 that $D_{4}$ shows a number appearing on one of $D_{1}, D_{2}$ and $D_{3}$ is