Three fives dice toss If four dice are tossed, find the probability that exactly 3 fives will show ( answer to the nearest thousandth in the for 0.xxx)?

Imagine that the dice are coloured A, B, C, D. Record the result of the rolling as a quadruple $(a,b,c,d)$, where $a$ is the number on the $A$-coloured die, $b$ is the number on the B-coloured die, and so on.
There are $6^4$ possible records, all equally likely.  
Now we count the number of records that have exactly three $5$'s. The location of the non-$5$ can be chosen in $4$ ways. For each such way, the value of the non-$5$ can be chosen in $5$ ways, for a total of $(4)(5)$. 
Thus our probability is $\frac{(4)(5)}{6^4}$.
Another way: (Well, it is not fundamentally different.) When we roll a die, call getting a $5$ a success, and getting a non-$5$ a failure. The probability of success on a single roll is $\frac{1}{6}$, and the probability of failure is $\frac{5}{6}$. 
If $X$ is the number of successes in $4$ trials, then $X$ has a Binomial Distribution. By a formula you may have seen already, 
$$\Pr(X=3)=\binom{4}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^1.$$
