The total number of outcomes is then $6^5$. There are $\binom{5}{3}$ ways to pick three dice that roll a two. The remaining two dice can roll any of the six faces. $$ \frac{\binom{5}{3}\times 6^2}{6^5} \approx 0.05 $$
I don't think this is right. Since if I try the above with 15 dice, the probability becomes greater than one.
On
I don't know if your question requires explicitly the use of combinatorics, but another, more direct way, to tackle the problem is with Random Variables.
If $X$ represents the number of 2s we then have $X \sim\text{Bin}(5,1/6)$. Thus
$$ \mathbb{P}(\text{al teast threee 2s})=\mathbb{P}(X \ge 3) = \sum_{i=3}^{5}\mathbb{P}(X=i)= \sum_{i=3}^{5} {5\choose i}\bigg(\frac{1}{6}\bigg)^i\bigg(\frac{5}{6}\bigg)^{5-i} \approx 3.55 \% $$
On
Actually, I do not agree that binomial distribution "should be used".You can get rid of overcounting using generalized version of principle of inclusion and exclusion.
If $L_m$ denotes the number of elements of $S$ that satisfies at least $m$ of the $t$ conditions, then we have the following formula $$L_m=S_m - \binom{m}{m-1}S_{m+1}+\binom{m+1}{m-1}S_{m+2}-...+(-1)^{t-m}\binom{t-1}{m-1}S_{t}$$
In our question $m=3$, so $$L_3=S_3-3S_4+6S_5$$ where
$$S_3=\frac{\binom{5}{3}\times 6^2}{6^5}=\frac{360}{6^5}$$
$$S_4=\frac{\binom{5}{4}\times 6^1}{6^5}=\frac{30}{6^5}$$
$$S_5=\frac{\binom{5}{5}\times 6^0}{6^5}= \frac{1}{6^5}$$
$$L_3=\frac{360}{6^5}- \bigg( 3\times \frac{30}{6^5}\bigg) + \bigg(6 \times \frac{1}{6^5}\bigg)=\frac{276}{7776}=0.035493827...$$
As @lulu indicated in the comments, this problem can be solved using the binomial distribution. The probability of exactly $k$ successes in $n$ trials of an event with probability $p$ of success in each trial is $$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$ where $p^k$ is the probability of $k$ successes, $(1 - p)^{n - k}$ is the probability of $n - k$ failures, and $\binom{n}{k}$ is the number of ways exactly $k$ successes could occur in $n$ trials.
In this problem, we define a success as rolling a $2$. Since $5$ dice are rolled, $n = 5$. Since we are assuming each die is fair, $p = 1/6 \implies 1 - p = 5/6$. Since we want to find the probability of at least three successes, we must add the probabilities of three successes, four successes, and five successes. Hence, \begin{align*} \Pr(X \ge 3) & = \sum_{k = 3}^{5} \binom{n}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{5 - k}\\ & = \binom{5}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^2 + \binom{5}{4}\left(\frac{1}{6}\right)^4\left(\frac{5}{6}\right) + \binom{5}{5}\left(\frac{1}{6}\right)^5 \end{align*}
To elaborate on lulu's comment about overcounting, you are counting each outcome with four twos four times, once for each of the $\binom{4}{3}$ ways of designating three of the four twos as the three twos, and each outcome with five twos ten times, once for each of the $\binom{5}{3}$ ways of designating three of the five twos as the three twos. For instance, you count the outcome $(2, 2, 3, 2, 2)$ four times: $(\color{red}{2}, \color{red}{2}, 3, \color{red}{2}, 2)$, $(\color{red}{2}, \color{red}{2}, 3, 2, \color{red}{2})$, $(\color{red}{2}, 2, 3, \color{red}{2}, \color{red}{2})$, $(2, \color{red}{2}, 3, \color{red}{2}, \color{red}{2})$.
Notice that $$\binom{5}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^2 + \color{red}{\binom{4}{3}}\binom{5}{4}\left(\frac{1}{6}\right)^4\left(\frac{5}{6}\right) + \color{red}{\binom{5}{3}}\binom{5}{5}\left(\frac{1}{6}\right)^5= \color{red}{\frac{\binom{5}{3}6^2}{6^5}}$$