<s>[INST] Four fair dice $D_1,D_2,D_3,D_4$ are rolled simultaneously . The probability that $D_4$ shows a number appearing on one of $D_1,D_2,D_3$ is 
Four fair dice $D_1,D_2,D_3,D_4$ 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 $D_1,D_2,D_3$ is

If $D_4$ shows the same number as $D_1$, then number of outcomes is $6\times 6\times 6$, since each dice has 6 possibilities except the 4th dice. (I took the 1st and 4th dice as a single dice).
Similarly, if $D_4$ shows the same number as $D_2$, then number of outcomes is $6\times 6\times 6$ and so on.
$$\text{Probability}=\frac{3\times6\times6\times6}{6\times6\times6\times6}=\frac{1}{2}$$
But the answer given is $\frac{91}{216}$
 [/INST] When $\{D_1,D_2,D_3\}$ has


*

*three elements, just select 3 numbers from $\{1,2,3,4,5,6\}$ with no repetition and arrange them. Then choose the number of $D_4$. Thus, number of events is $\binom{6}{3}3! \cdot 3 = 360$. 

*two elements, $D_1$ and $D_2$ is same or different. If $D_1$ and $D_2$ are same, then $D_3$ must have numbers different with $D_1$ and $D_2$. If $D_1$ and $D_2$ are different, then $D_3$ must have numbers same with $D_1$ or $D_2$. Thus, number of events is $6\cdot 5 \cdot 2 + 6\cdot 5 \cdot 2\cdot 2=180$.

*one elements, $D_1,D_2,D_3$, and $D_4$ must equal, so number of events is $6$.


Therefore, the answer is
$$
\frac{360+180+6}{6^4}=\frac{91}{216}.
$$
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