A fair coin is tossed 12 times. Find the probability that two heads do not occur consectively.

Answer

To find the probability that two heads do not occur consecutively when a fair coin is tossed 12 times, we can follow these steps:

### Step 1: Calculate Total Outcomes
When a fair coin is tossed 12 times, each toss has 2 possible outcomes (Heads or Tails). Therefore, the total number of outcomes is given by:

\[ \text{Total Outcomes} = 2^{12} = 4096 \]

### Step 2: Define the Problem
We need to count the number of favorable outcomes where no two heads occur consecutively.

### Step 3: Use a Recursive Approach
Let \( a_n \) be the number of valid sequences of length \( n \) where no two heads are consecutive. We can derive a recurrence relation for \( a_n \):
- If the first toss is T (Tails), the remaining \( n-1 \) tosses can be any valid sequence of length \( n-1 \), which contributes \( a_{n-1} \).
- If the first toss is H (Heads), the second toss must be T (to avoid consecutive heads), and the remaining \( n-2 \) tosses can be any valid sequence of length \( n-2 \), which contributes \( a_{n-2} \).

Thus, the recurrence relation is:

\[ a_n = a_{n-1} + a_{n-2} \]

### Step 4: Base Cases
We need base cases to start our recursion:
- \( a_1 = 2 \) (Sequences: H, T)
- \( a_2 = 3 \) (Sequences: HT, TH, TT)

### Step 5: Calculate \( a_n \) for \( n = 12 \)
Using the recurrence relation, we can calculate \( a_n \) for \( n = 3 \) to \( n = 12 \):

\[ \begin{align*} a_3 & = a_2 + a_1 = 3 + 2 = 5 \\ a_4 & = a_3 + a_2 = 5 + 3 = 8 \\ a_5 & = a_4 + a_3 = 8 + 5 = 13 \\ a_6 & = a_5 + a_4 = 13 + 8 = 21 \\ a_7 & = a_6 + a_5 = 21 + 13 = 34 \\ a_8 & = a_7 + a_6 = 34 + 21 = 55 \\ a_9 & = a_8 + a_7 = 55 + 34 = 89 \\ a_{10} & = a_9 + a_8 = 89 + 55 = 144 \\ a_{11} & = a_{10} + a_9 = 144 + 89 = 233 \\ a_{12} & = a_{11} + a_{10} = 233 + 144 = 377 \\ \end{align*} \]

### Step 6: Calculate the Probability
The probability that no two heads occur consecutively is given by the ratio of favorable outcomes to total outcomes:

\[ \text{Probability} = \frac{a_{12}}{2^{12}} = \frac{377}{4096} \]

### Step 7: Simplify the Probability
The fraction \( \frac{377}{4096} \) is already in its simplest form.

### Final Answer
Thus, the probability that two heads do not occur consecutively when a fair coin is tossed 12 times is:

\[ \frac{377}{4096} \]