Step-by-Step Solution

1. Determine the Total Number of Outcomes We are rolling four fair six-sided dice ($D_1, D_2, D_3, D_4$). Each die has 6 possible outcomes (numbers 1 to 6). Since the rolls are independent, the total number of possible outcomes in the sample space is: $N_{total} = 6 \times 6 \times 6 \times 6 = 6^4 = 1296$

2. Define the Event of Interest Let $E$ be the event that the number shown on die $D_4$ also appears on at least one of the dice $D_1, D_2,$ or $D_3$.

3. Use the Complementary Event Approach Calculating the probability of $E$ directly involves the principle of inclusion-exclusion, which can be complex. A simpler approach is to calculate the probability of the complementary event, $E'$, and then use the formula $P(E) = 1 - P(E')$.

   The complementary event $E'$ is that the number shown on $D_4$ does not appear on any of the dice $D_1, D_2,$ or $D_3$.

4. Calculate the Number of Favorable Outcomes for the Complementary Event ($E'$) To count the number of outcomes favorable to $E'$, we can fix the outcome of $D_4$ and then count the possibilities for the other dice.

   • There are 6 possible outcomes for the die $D_4$. Let's say the number shown on $D_4$ is $k$, where $k \in \{1, 2, 3, 4, 5, 6\}$.
   • For the event $E'$ to occur, the number on die $D_1$ must not be $k$. This leaves 5 possible outcomes for $D_1$ (any number from the set $\{1, 2, 3, 4, 5, 6\} \setminus \{k\}$).
   • Similarly, the number on die $D_2$ must not be $k$. This also leaves 5 possible outcomes for $D_2$.
   • Likewise, the number on die $D_3$ must not be $k$. This leaves 5 possible outcomes for $D_3$.

   The total number of outcomes favorable to $E'$, denoted as $N_{E'}$, is the product of the number of choices for each die: $N_{E'} = (\text{choices for } D_1) \times (\text{choices for } D_2) \times (\text{choices for } D_3) \times (\text{choices for } D_4)$ $N_{E'} = 5 \times 5 \times 5 \times 6 = 5^3 \times 6 = 125 \times 6 = 750$

5. Calculate the Probability of the Complementary Event ($P(E')$) The probability of $E'$ is the ratio of its favorable outcomes to the total number of outcomes: $P(E') = \frac{N_{E'}}{N_{total}} = \frac{750}{1296}$

6. Calculate the Probability of the Original Event ($P(E)$) Now, we can find the probability of the event $E$ using the complement rule: $P(E) = 1 - P(E') = 1 - \frac{750}{1296}$ To subtract, we find a common denominator: $P(E) = \frac{1296 - 750}{1296} = \frac{546}{1296}$

7. Simplify the Resulting Fraction Both the numerator and the denominator are divisible by 6: $546 \div 6 = 91$ $1296 \div 6 = 216$ So, the simplified probability is: $P(E) = \frac{91}{216}$ The prime factorization of 91 is $7 \times 13$, and the prime factorization of 216 is $2^3 \times 3^3$. They share no common factors, so the fraction is in its simplest form.

8. Conclusion The probability that $D_4$ shows a number appearing on one of $D_1, D_2,$ and $D_3$ is $\frac{91}{216}$. This corresponds to option A.