In an experiment a coin is tossed 10 times.
Q. The probability of the event that "exactly four heads occur and occur alternately" is :

Options

A. `1-(4)/(2^(10))`
B. `1-(7)/(2^(10))`
C. `(4)/(2^(10))`
D. `(5)/(2^(10))`

Answer

To solve the problem of finding the probability of getting exactly four heads that occur alternately when a coin is tossed 10 times, we can follow these steps:

### Step 1: Understand the arrangement of heads and tails
We need to have exactly 4 heads (H) and 6 tails (T) in such a way that the heads occur alternately. This means that the arrangement must look like HTH... or THT... depending on where we start.

### Step 2: Determine the valid arrangements
1. If we start with a head (H), the arrangement will look like:
- H T H T H T H T T T
- In this case, we can place heads in the 1st, 3rd, 5th, and 7th positions.

2. If we start with a tail (T), the arrangement will look like:
- T H T H T H T H T T
- In this case, we can place heads in the 2nd, 4th, 6th, and 8th positions.

### Step 3: Count the valid arrangements
From the above arrangements, we can see that there are only two valid arrangements for 4 heads and 6 tails:
- H T H T H T H T T T
- T H T H T H T H T T

Thus, there are a total of **2 favorable arrangements**.

### Step 4: Calculate the total number of outcomes
When a coin is tossed 10 times, the total number of possible outcomes is given by:
\[ \text{Total outcomes} = 2^{10} = 1024 \]

### Step 5: Calculate the probability
The probability of getting exactly 4 heads that occur alternately is given by the formula:
\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} \]
Substituting the values we found:
\[ \text{Probability} = \frac{2}{1024} = \frac{1}{512} \]

### Final Answer
The probability of the event that "exactly four heads occur and occur alternately" is:
\[ \frac{1}{512} \]

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