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 $\frac{91}{216}$
 A: 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 
$$1-\left(\frac{5}{6}\right)^3$$
A: 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
$$
\frac{360+180+6}{6^4}=\frac{91}{216}.
$$
A: I solved this question using Principle of inclusion and exclusion.
The number of ways in which two dies show the same number $= \binom{3}{1}(6)^3$ because I have to chose which dice uout of $D_1,D_2,D_3$ should show the same number as $D_4$.
But I have counted the number of ways in which $3$ dice show the same number twice. So I have to subtract $\binom{3}{2}(6)^2$
Now, I have subtracted the case where all $4$ dice show the same number twice. So I add back $\binom{3}{3}(6)$
$$\text{Required probability} =\frac{\binom{3}{1}(6)^3- \binom{3}{2}(6)^2+\binom{3}{3}(6)}{6^4}=\frac{91}{216}$$
A: Fix the number on D4.  The probability of  the event that D1 has a different number then the one on D4 is $\frac{5}{6}$, and this probability is same for D2 and D3. Therefore the event in which D1 D2 D3 has a different number than the one on D4 has a probability $(\frac{5}{6})^3= \frac{125}{216}$. Hence the event that we are interested in has a probability $1- \frac{125}{216} = \frac{91}{216}$.
