The number of possible outcomes in a throw of n ordinary dice in which at least one of the dice shows an odd number is

Options

A. `6^(n)-1`
B. `3^(n)-1`
C. `6^(n)-3^(n)`
D. `6^(n)-5^(n)`

Explanation

To solve the problem of finding the number of possible outcomes in a throw of \( n \) ordinary dice in which at least one of the dice shows an odd number, we can follow these steps:

### Step 1: Understand the total outcomes when throwing \( n \) dice
When we throw one ordinary die, there are 6 possible outcomes: {1, 2, 3, 4, 5, 6}. Therefore, when we throw \( n \) dice, the total number of outcomes is given by:
\[ \text{Total outcomes} = 6^n \]

### Step 2: Calculate the outcomes where all dice show even numbers
The even numbers on a die are {2, 4, 6}, which gives us 3 even outcomes. If we want all \( n \) dice to show even numbers, the number of outcomes for this scenario is:
\[ \text{Outcomes with all even numbers} = 3^n \]

### Step 3: Use the complement principle
To find the number of outcomes where at least one die shows an odd number, we can use the complement principle. This means we subtract the number of outcomes where all dice show even numbers from the total outcomes:
\[ \text{Outcomes with at least one odd number} = \text{Total outcomes} - \text{Outcomes with all even numbers} \]
Substituting the values we found:
\[ \text{Outcomes with at least one odd number} = 6^n - 3^n \]

### Conclusion
Thus, the number of possible outcomes in a throw of \( n \) ordinary dice in which at least one of the dice shows an odd number is:
\[ \boxed{6^n - 3^n} \]

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