A coin is tossed three times. If X denotes the absolute difference between the number of heads and the number of tails, then P(X=1)=

Options

A. `(1)/(2)`
B. `(2)/(3)`
C. `(1)/(6)`
D. `(3)/(4)`

Explanation

To solve the problem, we need to find the probability \( P(X = 1) \), where \( X \) is the absolute difference between the number of heads and the number of tails when a coin is tossed three times.

### Step-by-step Solution:

1. **Determine the Sample Space**:
When a coin is tossed three times, the total number of outcomes is \( 2^3 = 8 \). The sample space consists of the following outcomes:
- HHH
- HHT
- HTH
- THH
- HTT
- THT
- TTH
- TTT

2. **Count Heads and Tails**:
For each outcome, we can count the number of heads (H) and tails (T):
- HHH: 3 heads, 0 tails → \( |3 - 0| = 3 \)
- HHT: 2 heads, 1 tail → \( |2 - 1| = 1 \)
- HTH: 2 heads, 1 tail → \( |2 - 1| = 1 \)
- THH: 2 heads, 1 tail → \( |2 - 1| = 1 \)
- HTT: 1 head, 2 tails → \( |1 - 2| = 1 \)
- THT: 1 head, 2 tails → \( |1 - 2| = 1 \)
- TTH: 1 head, 2 tails → \( |1 - 2| = 1 \)
- TTT: 0 heads, 3 tails → \( |0 - 3| = 3 \)

3. **Identify Outcomes Where \( X = 1 \)**:
From the above calculations, the outcomes where \( X = 1 \) are:
- HHT
- HTH
- THH
- HTT
- THT
- TTH

Thus, there are 6 outcomes where \( X = 1 \).

4. **Calculate the Probability**:
The probability \( P(X = 1) \) is given by the ratio of the number of favorable outcomes to the total number of outcomes:
\[ P(X = 1) = \frac{\text{Number of outcomes where } X = 1}{\text{Total number of outcomes}} = \frac{6}{8} = \frac{3}{4} \]

### Final Answer:
\[ P(X = 1) = \frac{3}{4} \]