A fair 6 sided dice is rolled 4 times. What is the probability that at least 3 of the numbers will be either 1 or 6? I'd really love a sanity check here as I walk through what I believe is the solution.
Total possible outcomes = $6^4 = 1296$
Possible combinations of 3 rolls being either 1 or 6 = $({}_4C_3)\cdot2 = (4)\cdot2 = 8$
Also take into account all 1's and all 6's = $1 + 1 = 2$
Answer = $\frac{8+2}{ 1296} = \frac{10}{1296} = \mathbf{\frac{5}{648}} $
Really appreciate the help! :)
 A: I would use a binomial probability:
$$ \begin{align*}
P(\text{at least 3 are 1 or 6}) &= P(\text{exactly 3 are 1 or 6}) + P(\text{exactly 4 are 1 or 6}) \\
&= {}_4C_3 \left(\dfrac{2}{6}\right)^3\left(\dfrac{4}{6}\right)^1 + {}_4C_4 \left(\dfrac{2}{6}\right)^4\left(\dfrac{4}{6}\right)^0 \\
&= 4 \left(\dfrac{1}{3}\right)^3\left(\dfrac{2}{3}\right) + \left(\dfrac{1}{3}\right)^4 \\
&= \dfrac{8+1}{81} \\
&= \dfrac{1}{9} \\
\end{align*} $$
A: Your sample space of $1296$ can be replaced by a far smaller and more convenient sample space.  
But as an exercise we carry out the count of how many of the ordered quadruples satisfy the requirement of at least three $1$ and/or $6$.
First we count the number of cases where every entry is one of $1$ or $6$. There are plenty of choices other than $1111$ and $6666$, like $6111$. The first toss can take on anyone of $2$ values, and for each such value, there are $2$ possibilities for the second toss, and so on for a total of $2^4$. 
Next we count the number of cases where we have exactly three $1$ and/or $6$.
Where the oddball throw occurs can be chosen in $\binom{4}{1}$ ways. For each of these ways, the number on the oddball throw can be chosen in $4$ ways. And now the remaining three slots can be filled with $1$ or $6$ in $2^3$ ways, for a total of $128$.  
Thus $16+128$ elements of our sample space of $1296$ are "favourable." Now we can write down the probability.
A: $1$ or $6$ has probability $\cfrac 13$
All four $1$ or $6$ has probability $\left(\cfrac 13\right)^4=\cfrac 1{81}$
Precisely three $1$ or $6$ has probability (choosing a place out of four for the "other" value) $\dbinom 41\cfrac 23\left(\cfrac 13\right)^3=\cfrac 8{81}$
Add the two to get $\cfrac 9{81}=\cfrac 19$
A: To the top answer one must also add the possible combination of result with the value of remaining die which can be any value. So your total favourable outcome becomes 10 X 6 = 60. So actual probability of getting either 1 or 6 is 60/1296 = 5/108
