A biased coin with probability `p(0ltplt1)` of getting heads, is tossed until a head appears for the first time. If the probability that the number of tosses required is even is `1//5`, then p is equal to…….

Options

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

Explanation

To solve the problem, we need to find the probability \( p \) of getting heads when a biased coin is tossed until a head appears for the first time, given that the probability that the number of tosses required is even is \( \frac{1}{5} \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We are tossing a biased coin with probability \( p \) of landing heads.
- We want to find the probability that the first head appears on an even number of tosses.

2. **Identifying the Events**:
- The first head can appear on the 2nd toss, 4th toss, 6th toss, etc.
- The probability of getting the first head on the \( n \)-th toss is given by:
\[ P(\text{first head on } n\text{-th toss}) = (1-p)^{n-1} \cdot p \]
- For \( n \) to be even, we can express this as:
\[ P(\text{first head on even toss}) = \sum_{k=1}^{\infty} (1-p)^{2k-1} \cdot p \]
- This series represents the probability of getting the first head on the 2nd, 4th, 6th, etc.

3. **Summing the Series**:
- The series can be rewritten as:
\[ P(\text{even}) = p \sum_{k=1}^{\infty} (1-p)^{2k-1} \]
- The sum \( \sum_{k=1}^{\infty} (1-p)^{2k-1} \) is a geometric series with first term \( (1-p) \) and common ratio \( (1-p)^2 \):
\[ \sum_{k=1}^{\infty} (1-p)^{2k-1} = \frac{(1-p)}{1 - (1-p)^2} = \frac{(1-p)}{p(2-p)} \]

4. **Combining the Results**:
- Thus, we have:
\[ P(\text{even}) = p \cdot \frac{(1-p)}{p(2-p)} = \frac{1-p}{2-p} \]

5. **Setting Up the Equation**:
- We know from the problem statement that:
\[ \frac{1-p}{2-p} = \frac{1}{5} \]

6. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ 5(1-p) = 2-p \]
- Expanding and rearranging leads to:
\[ 5 - 5p = 2 - p \implies 5 - 2 = 5p - p \implies 3 = 4p \]

7. **Solving for \( p \)**:
- Dividing both sides by 4 gives:
\[ p = \frac{3}{4} \]

### Final Answer:
Thus, the probability \( p \) is \( \frac{3}{4} \).