A fair coin is tossed `9` times the probability that at least `5` consecutive heads occurs is

Answer

To find the probability that at least 5 consecutive heads occur when a fair coin is tossed 9 times, we can break the problem down into several cases based on the number of consecutive heads.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find the probability of getting at least 5 consecutive heads in 9 tosses of a fair coin. This means we need to consider cases where we have exactly 5, 6, 7, 8, or 9 consecutive heads.

2. **Total Outcomes**:
The total number of outcomes when a coin is tossed 9 times is \(2^9 = 512\).

3. **Case 1: Exactly 5 Consecutive Heads**:
- The 5 consecutive heads can start at different positions:
- Starting at position 1: HHHHHT___ (2 remaining positions can be filled with H or T)
- Starting at position 2: _HHHHHT__ (2 remaining positions can be filled with H or T)
- Starting at position 3: __HHHHHT_ (2 remaining positions can be filled with H or T)
- Starting at position 4: ___HHHHHT (2 remaining positions can be filled with H or T)
- Starting at position 5: ____HHHHH (2 remaining positions can be filled with H or T)
- Each of these configurations allows for \(2^3 = 8\) combinations for the remaining positions.
- Total for exactly 5 consecutive heads = \(5 \times 8 = 40\).

4. **Case 2: Exactly 6 Consecutive Heads**:
- The 6 consecutive heads can start at:
- Starting at position 1: HHHHHH___ (3 remaining positions can be filled with H or T)
- Starting at position 2: _HHHHHH__ (3 remaining positions can be filled with H or T)
- Starting at position 3: __HHHHHH_ (3 remaining positions can be filled with H or T)
- Starting at position 4: ___HHHHHH (3 remaining positions can be filled with H or T)
- Each configuration allows for \(2^3 = 8\) combinations.
- Total for exactly 6 consecutive heads = \(4 \times 8 = 32\).

5. **Case 3: Exactly 7 Consecutive Heads**:
- The 7 consecutive heads can start at:
- Starting at position 1: HHHHHHH__ (2 remaining positions can be filled with H or T)
- Starting at position 2: _HHHHHHH_ (2 remaining positions can be filled with H or T)
- Starting at position 3: __HHHHHHH (2 remaining positions can be filled with H or T)
- Each configuration allows for \(2^2 = 4\) combinations.
- Total for exactly 7 consecutive heads = \(3 \times 4 = 12\).

6. **Case 4: Exactly 8 Consecutive Heads**:
- The 8 consecutive heads can start at:
- Starting at position 1: HHHHHHHH_ (1 remaining position can be filled with H or T)
- Starting at position 2: _HHHHHHHH (1 remaining position can be filled with H or T)
- Each configuration allows for \(2^1 = 2\) combinations.
- Total for exactly 8 consecutive heads = \(2 \times 2 = 4\).

7. **Case 5: Exactly 9 Consecutive Heads**:
- There is only 1 way to have 9 consecutive heads: HHHHHHHHH.
- Total for exactly 9 consecutive heads = \(1\).

8. **Total Favorable Outcomes**:
- Adding all the cases together:
- Exactly 5 heads: 40
- Exactly 6 heads: 32
- Exactly 7 heads: 12
- Exactly 8 heads: 4
- Exactly 9 heads: 1
- Total = \(40 + 32 + 12 + 4 + 1 = 89\).

9. **Calculating the Probability**:
- The probability \(P\) of getting at least 5 consecutive heads is given by:
\[ P = \frac{\text{Total favorable outcomes}}{\text{Total outcomes}} = \frac{89}{512}. \]

### Final Answer:
The probability that at least 5 consecutive heads occur when a fair coin is tossed 9 times is \(\frac{89}{512}\).