A coin a tossed twice, that is the probability that at least one tail occurs?

Explanation

To solve the problem of finding the probability of getting at least one tail when a coin is tossed twice, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Sample Space**:
When a coin is tossed twice, the possible outcomes can be listed as:
- HH (Head, Head)
- HT (Head, Tail)
- TH (Tail, Head)
- TT (Tail, Tail)

Therefore, the sample space \( S \) is:
\[ S = \{ HH, HT, TH, TT \} \]

2. **Count the Total Outcomes**:
The total number of outcomes in the sample space is 4 (as we have listed 4 outcomes).

3. **Identify Favorable Outcomes**:
We need to find the outcomes that have at least one tail. The favorable outcomes are:
- HT
- TH
- TT

Thus, the favorable outcomes are:
\[ F = \{ HT, TH, TT \} \]

4. **Count the Favorable Outcomes**:
The number of favorable outcomes is 3 (HT, TH, TT).

5. **Calculate the Probability**:
The probability \( P \) of getting at least one tail can be calculated using the formula:
\[ P(\text{at least one tail}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{4} \]

6. **Final Answer**:
Therefore, the probability that at least one tail occurs when a coin is tossed twice is:
\[ P = \frac{3}{4} \]