A fair coin is tossed repeatedly. The probability of getting a result in fifth toss different from those obtained in the first four tosses is

Options

A. `(1)/(2)`
B. `(1)/(32)`
C. `(31)/(32)`
D. `(1)/(16)`

Explanation

To solve the problem, we need to determine the probability that the result of the fifth toss of a fair coin is different from the results of the first four tosses.

### Step-by-Step Solution:

1. **Understanding the Coin Toss**:
- A fair coin has two possible outcomes: Heads (H) or Tails (T).
- The probability of getting Heads (P(H)) = 1/2 and the probability of getting Tails (P(T)) = 1/2 for each toss.

2. **Identifying the Cases**:
- We need the fifth toss to be different from the first four tosses. This can happen in two scenarios:
- Case 1: The first four tosses are Tails (TTTT), and the fifth toss is Heads (H).
- Case 2: The first four tosses are Heads (HHHH), and the fifth toss is Tails (T).

3. **Calculating the Probability for Each Case**:
- **Case 1**: Probability of getting Tails in the first four tosses and Heads in the fifth toss:
\[ P(\text{TTTT}) = P(T) \times P(T) \times P(T) \times P(T) \times P(H) = \left(\frac{1}{2}\right)^4 \times \left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^5 = \frac{1}{32} \]
- **Case 2**: Probability of getting Heads in the first four tosses and Tails in the fifth toss:
\[ P(\text{HHHH}) = P(H) \times P(H) \times P(H) \times P(H) \times P(T) = \left(\frac{1}{2}\right)^4 \times \left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^5 = \frac{1}{32} \]

4. **Total Probability**:
- Since these two cases are mutually exclusive, we can add their probabilities to find the total probability of the fifth toss being different:
\[ P(\text{different}) = P(\text{Case 1}) + P(\text{Case 2}) = \frac{1}{32} + \frac{1}{32} = \frac{2}{32} = \frac{1}{16} \]

5. **Final Answer**:
- The probability that the result of the fifth toss is different from those obtained in the first four tosses is:
\[ \boxed{\frac{1}{16}} \]