A biased coin with probability `p(0 lt plt 1)` of falling tails is tossed until a tail appears for the first time. If the probability that tail comes in odd number of trials is `(2)/(3)`, then `p` equals

Options

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

Explanation

To solve the problem, we need to find the probability \( p \) of getting tails when a biased coin is tossed until a tail appears for the first time, given that the probability of getting tails in an odd number of trials is \( \frac{2}{3} \).

### Step-by-step Solution:

1. **Understanding the Problem**:
We know that the coin is biased, with the probability of tails being \( p \) and the probability of heads being \( 1 - p \). We want to find the probability \( p \) such that the probability of getting tails in an odd number of trials is \( \frac{2}{3} \).

2. **Identifying the Odd Trials**:
The odd trials where tails can appear for the first time are:
- 1st trial: Tails (probability = \( p \))
- 3rd trial: Heads in the first two trials and tails in the third (probability = \( (1 - p)^2 \cdot p \))
- 5th trial: Heads in the first four trials and tails in the fifth (probability = \( (1 - p)^4 \cdot p \))
- And so on...

3. **Setting Up the Probability**:
The total probability of getting tails in an odd number of trials can be expressed as an infinite series:
\[ P(\text{odd}) = p + (1 - p)^2 p + (1 - p)^4 p + \ldots \]
This can be factored as:
\[ P(\text{odd}) = p \left( 1 + (1 - p)^2 + (1 - p)^4 + \ldots \right) \]

4. **Recognizing the Series**:
The series \( 1 + (1 - p)^2 + (1 - p)^4 + \ldots \) is a geometric series with the first term \( a = 1 \) and common ratio \( r = (1 - p)^2 \). The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} = \frac{1}{1 - (1 - p)^2} \]

5. **Calculating the Sum**:
Therefore, we can write:
\[ P(\text{odd}) = p \cdot \frac{1}{1 - (1 - p)^2} \]
Simplifying the denominator:
\[ 1 - (1 - p)^2 = 1 - (1 - 2p + p^2) = 2p - p^2 \]
Thus,
\[ P(\text{odd}) = \frac{p}{2p - p^2} \]

6. **Setting Up the Equation**:
According to the problem, this probability equals \( \frac{2}{3} \):
\[ \frac{p}{2p - p^2} = \frac{2}{3} \]

7. **Cross Multiplying**:
Cross-multiplying gives:
\[ 3p = 2(2p - p^2) \]
Expanding the right side:
\[ 3p = 4p - 2p^2 \]

8. **Rearranging the Equation**:
Rearranging gives us:
\[ 2p^2 - p = 0 \]
Factoring out \( p \):
\[ p(2p - 1) = 0 \]

9. **Finding the Values of \( p \)**:
This gives us two solutions:
\[ p = 0 \quad \text{or} \quad 2p - 1 = 0 \implies p = \frac{1}{2} \]
Since \( p \) must be between 0 and 1, we have:
\[ p = \frac{1}{2} \]

### Final Answer:
The value of \( p \) is \( \frac{1}{2} \).