An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is :

Options

A. `(63)/(64)`
B. `(255)/(256)`
C. `(127)/(128)`
D. `(1)/(2)`

Answer

To solve the problem of finding the probability of obtaining at least one head and at least one tail when an unbiased coin is tossed eight times, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Total Outcomes**:
When a coin is tossed 8 times, each toss has 2 possible outcomes (Head or Tail). Therefore, the total number of outcomes when tossing the coin 8 times is:
\[ 2^8 = 256 \]

2. **Calculate the Probability of All Heads**:
The probability of getting heads in a single toss is \( \frac{1}{2} \). Therefore, the probability of getting all heads in 8 tosses is:
\[ P(\text{All Heads}) = \left(\frac{1}{2}\right)^8 = \frac{1}{256} \]

3. **Calculate the Probability of All Tails**:
Similarly, the probability of getting all tails in 8 tosses is also:
\[ P(\text{All Tails}) = \left(\frac{1}{2}\right)^8 = \frac{1}{256} \]

4. **Calculate the Probability of Getting All Heads or All Tails**:
Since these two events (all heads and all tails) are mutually exclusive, we can add their probabilities:
\[ P(\text{All Heads or All Tails}) = P(\text{All Heads}) + P(\text{All Tails}) = \frac{1}{256} + \frac{1}{256} = \frac{2}{256} = \frac{1}{128} \]

5. **Calculate the Probability of At Least One Head and One Tail**:
The probability of getting at least one head and at least one tail is the complement of the probability of getting all heads or all tails:
\[ P(\text{At least one head and one tail}) = 1 - P(\text{All Heads or All Tails}) = 1 - \frac{1}{128} = \frac{128 - 1}{128} = \frac{127}{128} \]

### Final Answer:
Thus, the probability of obtaining at least one head and at least one tail when an unbiased coin is tossed eight times is:
\[ \frac{127}{128} \]