When three standard dice are tossed, the numbers $a,b,c$ are obtained. Find the probability that $$(a-1)(b-1)(c-1) \neq 0$$
To find the probability of an event, I need to divide the number of favorable outcomes by the total number of possible outcomes. The total number of possible outcomes is $6^3 = 216$, since each die can show any of the six numbers independently. The favorable outcomes are those where none of the dice show a 1, since that would make the product zero. The probability of not getting a 1 on one die is $\frac{5}{6}$, and the probability of not getting a 1 on all three dice is $\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{125}{216}$. Therefore, the probability that $(a-1)(b-1)(c-1) \neq 0$ is $\frac{125}{216}$. 