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 $\frac{91}{216}$
Best Answer
Think in terms of the complementary event. Suppose $a$ shows on $D_4$, then what is the probability that none of $D_i$'s ($i \in \{1,2,3\}$) have $a$.
The probability of this event is $\left(\frac{5}{6}\right)^3$. So the probability of event you want is $$1-\left(\frac{5}{6}\right)^3$$