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)?
 A: HINT: Imagine rolling the dice one at a time. Say that a set of four rolls is good if exactly $3$ of the $4$ are fives. Then there are $4$ kinds of good roll: the non-five can come on the first die, the second, the third, or the fourth. Suppose, for example, that it comes on the second roll; let’s calculate the probability of this outcome.


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*The first roll was a five; that occurs with probability $\frac16$. The second roll was not a five; that occurs with probability $1-\frac16$. The third roll was a five again: probability $\frac16$. And the fourth roll was also a five: again probability $\frac16$. How do you combine these numbers to get the probability of a good roll with the non-fine coming on the second die?

*What are the probabilities of the other three kinds of good sets of rolls?

*How do you combine these probabilities to get the total probability of a good set of rolls?
A: 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, 
$$\Pr(X=3)=\binom{4}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^1.$$
