A biased coin with probability p, `0ltplt1` of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is 2/5, then p equals

Options

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

Explanation

To solve the problem, we need to find the probability \( p \) such that the probability of getting an even number of tosses before getting a head is \( \frac{2}{5} \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
We are tossing a biased coin until we get a head. The probability of getting heads is \( p \) and the probability of getting tails is \( 1 - p \). We need to find the probability \( p \) such that the probability of the number of tosses being even is \( \frac{2}{5} \).

2. **Identifying Even Toss Outcomes**:
The outcomes where we get heads on an even toss can occur in the following ways:
- Getting a head on the 2nd toss: This means we get tails on the 1st toss and heads on the 2nd toss.
- Getting a head on the 4th toss: This means we get tails on the 1st, 2nd, and 3rd tosses, and heads on the 4th toss.
- This pattern continues for all even-numbered tosses.

3. **Calculating the Probability for Even Tosses**:
The probability of getting heads on the 2nd toss is:
\[ (1 - p) \cdot p \]
The probability of getting heads on the 4th toss is:
\[ (1 - p)^3 \cdot p \]
The probability of getting heads on the 6th toss is:
\[ (1 - p)^5 \cdot p \]
Thus, the total probability of getting heads on an even toss is:
\[ P(\text{even}) = (1 - p)p + (1 - p)^3 p + (1 - p)^5 p + \ldots \]

4. **Factoring Out \( p \)**:
We can factor out \( p \):
\[ P(\text{even}) = p \left[ (1 - p) + (1 - p)^3 + (1 - p)^5 + \ldots \right] \]

5. **Recognizing the Series**:
The series inside the brackets is a geometric series with the first term \( (1 - p) \) and common ratio \( (1 - p)^2 \):
\[ S = (1 - p) + (1 - p)^3 + (1 - p)^5 + \ldots \]
The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio. Here, \( a = (1 - p) \) and \( r = (1 - p)^2 \):
\[ S = \frac{(1 - p)}{1 - (1 - p)^2} = \frac{(1 - p)}{p(2 - p)} \]

6. **Putting It All Together**:
Thus, we have:
\[ P(\text{even}) = p \cdot \frac{(1 - p)}{p(2 - p)} = \frac{(1 - p)}{(2 - p)} \]
We know from the problem statement that:
\[ \frac{(1 - p)}{(2 - p)} = \frac{2}{5} \]

7. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 5(1 - p) = 2(2 - p) \]
Expanding both sides:
\[ 5 - 5p = 4 - 2p \]

8. **Rearranging the Equation**:
Rearranging gives:
\[ 5 - 4 = 5p - 2p \implies 1 = 3p \implies p = \frac{1}{3} \]

### Final Answer:
Thus, the value of \( p \) is:
\[ \boxed{\frac{1}{3}} \]