When three standard dice are tossed, the numbers $a,b,c$ are obtained. Find the probability that $$(a-1)(b-1)(c-1) \neq 0$$
None of $a,b,c$ are equal to 1 if and only if $(a-1)(b-1)(c-1) \neq 0$. $a,b,c$ can only be $2,3,4,5,6$. The probability of this is $\frac{5^3}{6^3}=\boxed{\frac{125}{216}}$.
\frac{125}{216}
Level 3
Counting & Probability
Problem: When three standard dice are tossed, the numbers $a,b,c$ are obtained. Find the probability that $$(a-1)(b-1)(c-1) \neq 0$$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the probability that \((a-1)(b-1)(c-1) \neq 0\), we first interpret the condition. The expression \((a-1)(b-1)(c-1) \neq 0\) implies that none of the variables \(a\), \(b\), or \(c\) can equal 1. Therefore, we need to determine the probability that all three dice show numbers from the set \{2, 3, 4, 5, 6\}.

Each die has 6 faces, and the total number of outcomes when rolling three dice is \(6^3 = 216\).

Next, we calculate the number of favorable outcomes where \(a\), \(b\), and \(c\) are all greater than or equal to 2. The possible outcomes for each die are now reduced to 5 options: \{2, 3, 4, 5, 6\}. Thus, the number of favorable outcomes is \(5^3 = 125\).

Now, we can find the probability by dividing the number of favorable outcomes by the total number of outcomes:

\[
P((a-1)(b-1)(c-1) \neq 0) = \frac{125}{216}
\]

This calculation confirms that the probability that none of the dice show a 1 is indeed \(\frac{125}{216}\).<|END_OF_LATENT|> The final answer is $\frac{125}{216}$.