FORMULABANK

First, we determine the probability that none of the 3 throws are sixes.

P(no six) = $\frac{5·5·5}{6·6·6}=\frac{125}{216}$

This means that 125 of the 216 different combinations that are possible when rolling a standard die three times do NOT contain one or more sixes. Therefore, there must be $(216-125)$ = $91$ combinations that contain at least one six.

Therefore, the probability of rolling at least one six in three throws is

P(at least one six) = $\frac{216}{216}-\frac{125}{216}=\frac{91}{216}\approx 0.423\approx 42.3\%$