Let n ordinary fair dice are rolled once. The probability that at least one of the dice shows an odd number is `((31)/(32))` than 'n' is equal to :

Options

A. 3
B. 4
C. 5
D. 6

Answer

To solve the problem, we need to find the value of \( n \) such that the probability of rolling at least one odd number with \( n \) ordinary fair dice is \( \frac{31}{32} \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find the probability that at least one die shows an odd number when \( n \) dice are rolled.

2. **Identify the Probability of Success and Failure**:
When a single die is rolled, the outcomes are 1, 2, 3, 4, 5, and 6. The odd numbers are 1, 3, and 5.
- Probability of rolling an odd number (success) \( p = \frac{3}{6} = \frac{1}{2} \).
- Probability of rolling an even number (failure) \( q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2} \).

3. **Using the Complement Rule**:
The probability of getting at least one odd number when rolling \( n \) dice can be calculated using the complement:
\[ P(\text{at least one odd}) = 1 - P(\text{no odd numbers}) \]
Therefore, we can express it as:
\[ P(\text{at least one odd}) = 1 - P(X = 0) \]

4. **Calculating \( P(X = 0) \)**:
The probability of getting no odd numbers (i.e., all dice show even numbers) when rolling \( n \) dice is:
\[ P(X = 0) = q^n = \left(\frac{1}{2}\right)^n \]
Thus, we have:
\[ P(\text{at least one odd}) = 1 - \left(\frac{1}{2}\right)^n \]

5. **Setting Up the Equation**:
According to the problem, this probability is given as \( \frac{31}{32} \):
\[ 1 - \left(\frac{1}{2}\right)^n = \frac{31}{32} \]

6. **Solving for \( n \)**:
Rearranging the equation gives:
\[ \left(\frac{1}{2}\right)^n = 1 - \frac{31}{32} = \frac{1}{32} \]
This can be rewritten as:
\[ \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^5 \]
By comparing the exponents, we find:
\[ n = 5 \]

### Conclusion:
Thus, the value of \( n \) is \( 5 \).