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 of finding the probability that the result of the fifth toss of a fair coin is different from the results obtained in the first four tosses, we can follow these steps:

### Step 1: Understand the Outcomes of the First Four Tosses
When a fair coin is tossed, there are two possible outcomes for each toss: Heads (H) or Tails (T). In the first four tosses, we can have any combination of H and T.

### Step 2: Determine the Possible Outcomes for the Fifth Toss
The fifth toss must be different from the results of the first four tosses. This means if the first four tosses resulted in all Heads (HHHH), the fifth toss must be Tails (T). Conversely, if the first four tosses resulted in all Tails (TTTT), the fifth toss must be Heads (H).

### Step 3: Calculate the Total Outcomes for the First Four Tosses
For four tosses, since each toss has 2 outcomes, the total number of outcomes for the first four tosses is:
\[ 2^4 = 16 \]

### Step 4: Identify the Outcomes for the Fifth Toss
For the fifth toss to be different from the first four tosses, we can have two scenarios:
1. All four tosses are Heads (HHHH) and the fifth toss is Tails (T).
2. All four tosses are Tails (TTTT) and the fifth toss is Heads (H).

### Step 5: Calculate the Probability of Each Scenario
- The probability of getting all Heads in the first four tosses is:
\[ P(HHHH) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \]
- The probability of getting all Tails in the first four tosses is:
\[ P(TTTT) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \]

### Step 6: Combine the Probabilities
Since these two scenarios are mutually exclusive (they cannot happen at the same time), we can add their probabilities:
\[ P(\text{fifth toss different}) = P(HHHH) + P(TTTT) = \frac{1}{16} + \frac{1}{16} = \frac{2}{16} = \frac{1}{8} \]

### Conclusion
Thus, the probability that the result of the fifth toss is different from those obtained in the first four tosses is:
\[ \frac{1}{2} \]