A fair coin is tossed 10 times. Then the probability that two heads do not occur consecutively is `7//64` b. `1//8` c. `9//16` d. `9//64`

Options

A. `7//64`
B. `1//8`
C. `9//16`
D. `9//64`

Explanation

To solve the problem of finding the probability that two heads do not occur consecutively when a fair coin is tossed 10 times, we can use combinatorial methods. Here’s a step-by-step solution:

### Step 1: Define the Problem
We need to find the total number of ways to toss a coin 10 times such that no two heads (H) are consecutive.

### Step 2: Use Recursion
Let \( a_n \) be the number of valid sequences of length \( n \) that do not have two consecutive heads. We can derive a recurrence relation:
- If the first toss is a tail (T), the remaining \( n-1 \) tosses can be any valid sequence of length \( n-1 \).
- If the first toss is a head (H), the second toss must be a tail (T) to avoid consecutive heads, leaving \( n-2 \) tosses that can be any valid sequence of length \( n-2 \).

Thus, the recurrence relation is:
\[ a_n = a_{n-1} + a_{n-2} \]

### Step 3: Base Cases
We need to establish the base cases:
- \( a_1 = 2 \) (sequences: H, T)
- \( a_2 = 3 \) (sequences: HT, TH, TT)

### Step 4: Calculate Values Up to \( n = 10 \)
Using the recurrence relation, we can calculate \( a_n \) for \( n = 3 \) to \( n = 10 \):
- \( a_3 = a_2 + a_1 = 3 + 2 = 5 \)
- \( a_4 = a_3 + a_2 = 5 + 3 = 8 \)
- \( a_5 = a_4 + a_3 = 8 + 5 = 13 \)
- \( a_6 = a_5 + a_4 = 13 + 8 = 21 \)
- \( a_7 = a_6 + a_5 = 21 + 13 = 34 \)
- \( a_8 = a_7 + a_6 = 34 + 21 = 55 \)
- \( a_9 = a_8 + a_7 = 55 + 34 = 89 \)
- \( a_{10} = a_9 + a_8 = 89 + 55 = 144 \)

### Step 5: Total Outcomes
The total number of outcomes when tossing a coin 10 times is \( 2^{10} = 1024 \).

### Step 6: Calculate the Probability
The probability that two heads do not occur consecutively is given by:
\[ P = \frac{a_{10}}{2^{10}} = \frac{144}{1024} = \frac{9}{64} \]

### Final Answer
Thus, the probability that two heads do not occur consecutively when a fair coin is tossed 10 times is:
\[ \boxed{\frac{9}{64}} \]

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