A fair coin is tossed 10 times. If the probability that heads never occur on consecutive tosses be `(m)/(n)` (where m, n are coprime and m, `n in N`), then the value of `(n-7m)` equals to :

Answer

To solve the problem of finding the probability that heads never occur on consecutive tosses when a fair coin is tossed 10 times, we can follow these steps:

### Step-by-Step Solution:

1. **Total Outcomes**:
When a coin is tossed 10 times, the total number of outcomes is given by \(2^{10}\).
\[ \text{Total outcomes} = 2^{10} = 1024 \]

**Hint**: Remember that each toss has 2 possible outcomes (heads or tails).

2. **Define Favorable Outcomes**:
We need to count the number of arrangements where heads (H) never occur consecutively. We can use a combinatorial approach to find the number of valid sequences.

3. **Using Fibonacci Sequence**:
Let \(a_n\) represent the number of valid sequences of length \(n\) where heads do not occur consecutively. The recurrence relation can be defined as:
\[ a_n = a_{n-1} + a_{n-2} \]
This is because:
- If the last toss is a tail (T), the first \(n-1\) can be any valid sequence of length \(n-1\).
- If the last toss is a head (H), the second last must be a tail, and the first \(n-2\) can be any valid sequence of length \(n-2\).

4. **Base Cases**:
We need base cases to start our recurrence:
- \(a_1 = 2\) (H, T)
- \(a_2 = 3\) (HT, TH, TT)

5. **Calculate \(a_n\) for \(n = 10\)**:
Using the recurrence relation:
\[ \begin{align*} a_3 & = a_2 + a_1 = 3 + 2 = 5 \\ a_4 & = a_3 + a_2 = 5 + 3 = 8 \\ a_5 & = a_4 + a_3 = 8 + 5 = 13 \\ a_6 & = a_5 + a_4 = 13 + 8 = 21 \\ a_7 & = a_6 + a_5 = 21 + 13 = 34 \\ a_8 & = a_7 + a_6 = 34 + 21 = 55 \\ a_9 & = a_8 + a_7 = 55 + 34 = 89 \\ a_{10} & = a_9 + a_8 = 89 + 55 = 144 \end{align*} \]

6. **Calculate the Probability**:
The probability \(P\) that heads never occur consecutively is given by the ratio of favorable outcomes to total outcomes:
\[ P = \frac{a_{10}}{2^{10}} = \frac{144}{1024} = \frac{9}{64} \]

7. **Identify \(m\) and \(n\)**:
From the probability \(\frac{m}{n} = \frac{9}{64}\), we have \(m = 9\) and \(n = 64\).

8. **Calculate \(n - 7m\)**:
Now we need to find \(n - 7m\):
\[ n - 7m = 64 - 7 \times 9 = 64 - 63 = 1 \]

### Final Answer:
The value of \(n - 7m\) is \(1\).