A die is tossed thrice. Find the probability of getting an even number at least once

Answer

To find the probability of getting an even number at least once when a die is tossed three times, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Total Outcomes**:
A die has 6 faces numbered from 1 to 6. The even numbers on a die are 2, 4, and 6. Therefore, there are 3 even numbers and 3 odd numbers.

2. **Calculate the Probability of Getting an Odd Number**:
The probability of getting an odd number in a single toss of the die is:
\[ P(\text{Odd}) = \frac{\text{Number of Odd Outcomes}}{\text{Total Outcomes}} = \frac{3}{6} = \frac{1}{2} \]

3. **Calculate the Probability of Getting Odd Numbers in All Three Tosses**:
Since the tosses are independent, the probability of getting odd numbers in all three tosses is:
\[ P(\text{All Odd}) = P(\text{Odd}) \times P(\text{Odd}) \times P(\text{Odd}) = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \]

4. **Calculate the Probability of Getting at Least One Even Number**:
The probability of getting at least one even number is the complement of getting all odd numbers:
\[ P(\text{At least one Even}) = 1 - P(\text{All Odd}) = 1 - \frac{1}{8} = \frac{7}{8} \]

### Final Answer:
Thus, the probability of getting an even number at least once when a die is tossed three times is:
\[ \frac{7}{8} \]