Probability of Fourth Die Showing Number Appearing on Other Three Dice

To find the probability that $D_4$ shows a number appearing on at least one of the dice $D_1, D_2,$ and $D_3$, we can compute the probability of the complementary event (i.e., $D_4$ shows a number that does not appear on any of $D_1, D_2,$ and $D_3$) and subtract it from $1$.

Let the total number of outcomes when four fair dice are rolled simultaneously be $N$: $N = 6 \times 6 \times 6 \times 6 = 6^4 = 1296$

Now, let us calculate the number of outcomes where the number shown on $D_4$ does not appear on $D_1, D_2,$ or $D_3$:

1. $D_4$ can show any of the $6$ possible numbers $\{1, 2, 3, 4, 5, 6\}$.
2. For a fixed outcome on $D_4$, each of the dice $D_1, D_2,$ and $D_3$ must show any number other than the one shown on $D_4$. Hence, there are $5$ choices for each of $D_1, D_2,$ and $D_3$.

Thus, the number of unfavorable outcomes $n(E^c)$ is: $n(E^c) = 6 \times 5 \times 5 \times 5 = 6 \times 5^3 = 750$

The probability of the complementary event, $P(E^c)$, is: $P(E^c) = \frac{n(E^c)}{N} = \frac{6 \times 5^3}{6^4} = \left(\frac{5}{6}\right)^3 = \frac{125}{216}$

Therefore, the required probability $P(E)$ that $D_4$ shows a number appearing on one of $D_1, D_2,$ and $D_3$ is: $P(E) = 1 - P(E^c) = 1 - \frac{125}{216} = \frac{216 - 125}{216} = \frac{91}{216}$

Thus, the correct option is A.