jmc

There are $\left(\genfrac{}{}{0ex}{}{10}{3}\right)=120$ ways to select exactly three dice to roll 1's ones out of ten dice total. The probability of any one of these outcomes occurring is ${\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^7$ and they're all mutually exclusive cases, so the probability that one of them will occur (which is the probability that we're looking for) is $\left(\genfrac{}{}{0ex}{}{10}{3}\right){\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^7=\frac{120\cdot 5^7\cdot 1^3}{6^{10}}\approx \boxed{.155}$.