In an experiment a coin is tossed 10 times.
Q. Probability that no two heads are consecutive is :

Options

A. `(143)/(2^(10))`
B. `(9)/(2^(6))`
C. `(2^(7)-1)/(2^(10))`
D. `(2^(6)-1)/(2^(6))`

Answer

To find the probability that no two heads are consecutive when a coin is tossed 10 times, we can follow these steps:

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

### Step 2: Define the Problem
Let \( p_n \) be the number of valid sequences of heads (H) and tails (T) of length \( n \) such that no two heads are consecutive. We need to find \( p_{10} \).

### Step 3: Establish a Recurrence Relation
To form a valid sequence of length \( n \):
- If the first toss is T, the remaining \( n-1 \) tosses can be any valid sequence of length \( n-1 \) (i.e., \( p_{n-1} \)).
- If the first toss is H, the second toss must be T (to avoid consecutive heads), and the remaining \( n-2 \) tosses can be any valid sequence of length \( n-2 \) (i.e., \( p_{n-2} \)).

Thus, we can establish the recurrence relation:
\[ p_n = p_{n-1} + p_{n-2} \]

### Step 4: Base Cases
We need to define the base cases:
- \( p_1 = 2 \) (sequences: H, T)
- \( p_2 = 3 \) (sequences: HT, TH, TT)

### Step 5: Calculate \( p_n \) for \( n = 3 \) to \( n = 10 \)
Using the recurrence relation, we can compute the values:
- \( p_3 = p_2 + p_1 = 3 + 2 = 5 \)
- \( p_4 = p_3 + p_2 = 5 + 3 = 8 \)
- \( p_5 = p_4 + p_3 = 8 + 5 = 13 \)
- \( p_6 = p_5 + p_4 = 13 + 8 = 21 \)
- \( p_7 = p_6 + p_5 = 21 + 13 = 34 \)
- \( p_8 = p_7 + p_6 = 34 + 21 = 55 \)
- \( p_9 = p_8 + p_7 = 55 + 34 = 89 \)
- \( p_{10} = p_9 + p_8 = 89 + 55 = 144 \)

### Step 6: Calculate the Probability
Now, the probability that no two heads are consecutive is given by:
\[ P(\text{no two heads consecutive}) = \frac{p_{10}}{\text{Total Outcomes}} = \frac{144}{1024} \]
This simplifies to:
\[ P = \frac{144}{1024} = \frac{9}{64} \]

### Final Answer
Thus, the probability that no two heads are consecutive when a coin is tossed 10 times is:
\[ \boxed{\frac{9}{64}} \]