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
$\frac{91}{216}$

Explanation for the correct option:
Find the Probability :
Let $E$ be the event of the last one does not appear in the first three. and $A$ be the total number of outcomes
Given that Four fair dice $D_{1}, D_{2}, D_{3}$ and $D_{4}$ rolled simultaneously.
The total number of outcomes $n (A) = 6 \times 6 \times 6 \times 6$ .
Suppose, the outcome of the last one does not appear in the first three.
The number of such outcomes possible $n (E) = 6 \times 5^{3}$
( $∵ D_{4}$ can show any of $6$ numbers and the first three can show any of the remaining $5$ numbers)
Therefore, the probability that $D_{4}$ does not show a number shown by any of $D_{1}, D_{2}$ or $D_{3}$
$\frac{n (E)}{n (A)} = \frac{6 \times 5^{3}}{6^{4}}$ $[∵ P (E) = \frac{n (E)}{n (A)}]$
$= \frac{125}{216}$
So, the required probability is
$\begin{array}{rcl} P (E') & = & 1 - \frac{125}{216} \\ = & \frac{91}{216} \end{array}$ $[∵ P (E') = 1 - P (E)]$
Hence the correct option is A.

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