A coin is tossed 7 times. Then the probability that at least 4 consective heads apear is

Options

A. `3//16`
B. `5//32`
C. `3//16`
D. `1//8`

Explanation

To find the probability that at least 4 consecutive heads appear when a coin is tossed 7 times, we can approach the problem by breaking it down into cases based on the number of consecutive heads.

### Step-by-Step Solution:

1. **Understanding the Total Outcomes**:
- When a coin is tossed 7 times, the total number of outcomes is \(2^7 = 128\).

2. **Case 1: Exactly 4 Consecutive Heads**:
- We can have 4 consecutive heads in various positions:
- HHHHTxx (where x can be H or T)
- xHHHHTx
- xxHHHHH
- For each of these arrangements, the remaining positions (x) can be either H or T.
- The positions can be:
- HHHHTxx: 1 way (4 heads) + 2 choices for the last 2 positions = \(2^2 = 4\) outcomes
- xHHHHTx: 1 way (4 heads) + 1 choice for the first position + 1 choice for the last position = \(2^2 = 4\) outcomes
- xxHHHHH: 1 way (4 heads) + 2 choices for the first 2 positions = \(2^2 = 4\) outcomes
- Total for this case = \(4 + 4 + 4 = 12\) outcomes.

3. **Case 2: Exactly 5 Consecutive Heads**:
- Possible arrangements:
- HHHHHTx
- xHHHHHT
- xxHHHHH
- For each arrangement:
- HHHHHTx: 1 way (5 heads) + 1 choice for the last position = \(2^1 = 2\) outcomes
- xHHHHHT: 1 way (5 heads) + 1 choice for the first position = \(2^1 = 2\) outcomes
- xxHHHHH: 1 way (5 heads) + 2 choices for the first 2 positions = \(2^2 = 4\) outcomes
- Total for this case = \(2 + 2 + 4 = 8\) outcomes.

4. **Case 3: Exactly 6 Consecutive Heads**:
- Possible arrangements:
- HHHHHHx
- xHHHHHH
- For each arrangement:
- HHHHHHx: 1 way (6 heads) + 1 choice for the last position = \(2^1 = 2\) outcomes
- xHHHHHH: 1 way (6 heads) + 1 choice for the first position = \(2^1 = 2\) outcomes
- Total for this case = \(2 + 2 = 4\) outcomes.

5. **Case 4: Exactly 7 Consecutive Heads**:
- There is only 1 arrangement: HHHHHHH.
- Total for this case = 1 outcome.

6. **Calculating the Total Outcomes with At Least 4 Consecutive Heads**:
- Total outcomes = Case 1 + Case 2 + Case 3 + Case 4
- Total = \(12 + 8 + 4 + 1 = 25\) outcomes.

7. **Calculating the Probability**:
- The probability of getting at least 4 consecutive heads = \(\frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{25}{128}\).

### Final Answer:
The probability that at least 4 consecutive heads appear when a coin is tossed 7 times is \(\frac{25}{128}\).