A biased coin with probability `P,(0 lt p ,1)` of heads is tossed until a head appear for the first time. If the probability that the number of tosses required is even is `2/5` then `p=`

Answer

To solve the problem, we need to find the value of \( p \) given that the probability of getting heads on an even toss is \( \frac{2}{5} \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
We are tossing a biased coin until we get heads for the first time. The probability of getting heads is \( p \) and the probability of getting tails is \( 1 - p \). We need to find the probability that the first head appears on an even toss.

2. **Probability of Getting Heads on Even Tosses**:
The first head can appear on the 2nd, 4th, 6th, etc. toss. The probability of getting heads on the 2nd toss is:
\[ (1 - p) \cdot p \]
The probability of getting heads on the 4th toss is:
\[ (1 - p)^3 \cdot p \]
Similarly, for the 6th toss, it is:
\[ (1 - p)^5 \cdot p \]
Therefore, the total probability \( P(\text{even}) \) that the first head appears on an even toss is:
\[ P(\text{even}) = (1 - p)p + (1 - p)^3 p + (1 - p)^5 p + \ldots \]

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

4. **Recognizing the Geometric Series**:
The series inside the brackets is a geometric series where the first term \( a = (1 - p) \) and the common ratio \( r = (1 - p)^2 \):
\[ S = (1 - p) + (1 - p)^3 + (1 - p)^5 + \ldots = \frac{(1 - p)}{1 - (1 - p)^2} \]

5. **Calculating the Denominator**:
The denominator simplifies as follows:
\[ 1 - (1 - p)^2 = 1 - (1 - 2p + p^2) = 2p - p^2 \]

6. **Putting it All Together**:
Now substituting back into our expression for \( P(\text{even}) \):
\[ P(\text{even}) = p \cdot \frac{(1 - p)}{2p - p^2} \]

7. **Setting the Equation**:
We know that \( P(\text{even}) = \frac{2}{5} \):
\[ p \cdot \frac{(1 - p)}{2p - p^2} = \frac{2}{5} \]

8. **Cross Multiplying**:
Cross multiplying gives:
\[ 5p(1 - p) = 2(2p - p^2) \]

9. **Expanding Both Sides**:
Expanding both sides:
\[ 5p - 5p^2 = 4p - 2p^2 \]

10. **Rearranging the Equation**:
Rearranging gives:
\[ 5p^2 - 2p^2 - 5p + 4p = 0 \]
This simplifies to:
\[ 3p^2 - p = 0 \]

11. **Factoring Out \( p \)**:
Factoring out \( p \):
\[ p(3p - 1) = 0 \]

12. **Finding the Values of \( p \)**:
This gives us two solutions:
\[ p = 0 \quad \text{or} \quad 3p - 1 = 0 \Rightarrow p = \frac{1}{3} \]
Since \( p \) must be between 0 and 1, we have:
\[ p = \frac{1}{3} \]

### Final Answer:
\[ p = \frac{1}{3} \]