A biased coin with the probability p, 0

Let's break down the problem step by step.

We are given a biased coin with the probability of heads being \( p \) and the probability of tails being \( 1 - p \). We are tossing the coin until the first head appears, and the task is to find the value of \( p \) such that the probability of an even number of tosses is \( \frac{2}{5} \).

### Step 1: Understand the problem
The coin is tossed repeatedly until the first head appears, and we are interested in the probability that the number of tosses required is even. This means that the first head appears on the second toss, fourth toss, sixth toss, and so on.

### Step 2: Probability of an even number of tosses
The key to solving this problem is realizing that if the first head appears on the \( 2n \)-th toss, the sequence of outcomes must be:

- \( (n-1) \) tails followed by a head on the \( 2n \)-th toss.

Thus, the probability of this happening is:
\[
(1 - p)^{2n-1} \times p
\]
where \( (1 - p)^{2n-1} \) represents the \( 2n-1 \) tails before the head, and \( p \) represents the probability of getting a head on the \( 2n \)-th toss.

To find the total probability of getting an even number of tosses, we sum the probabilities for all even numbers of tosses:
\[
P(\text{even}) = \sum_{n=1}^{\infty} (1 - p)^{2n-1} \times p
\]

### Step 3: Simplifying the series
This is a geometric series with the first term \( p(1 - p) \) and common ratio \( (1 - p)^2 \). The sum of an infinite geometric series \( \sum_{n=0}^{\infty} ar^n \) is given by:
\[
S = \frac{a}{1 - r}
\]
where \( a \) is the first term and \( r \) is the common ratio. In our case:
- \( a = p(1 - p) \)
- \( r = (1 - p)^2 \)

Thus, the total probability is:
\[
P(\text{even}) = \frac{p(1 - p)}{1 - (1 - p)^2}
\]

### Step 4: Set up the equation
We are told that the probability of an even number of tosses is \( \frac{2}{5} \). So, we set:
\[
\frac{p(1 - p)}{1 - (1 - p)^2} = \frac{2}{5}
\]

### Step 5: Solve the equation
First, simplify the denominator:
\[
1 - (1 - p)^2 = 1 - (1 - 2p + p^2) = 2p - p^2
\]
So the equation becomes:
\[
\frac{p(1 - p)}{2p - p^2} = \frac{2}{5}
\]

Now, cross-multiply:
\[
5p(1 - p) = 2(2p - p^2)
\]
Expand both sides:
\[
5p - 5p^2 = 4p - 2p^2
\]
Simplify:
\[
5p - 4p = 5p^2 - 2p^2
\]
\[
p = 3p^2
\]

Solve for \( p \):
\[
3p^2 - p = 0
\]
\[
p(3p - 1) = 0
\]
Thus, \( p = 0 \) or \( p = \frac{1}{3} \). Since \( p = 0 \) is not a valid solution (as \( 0 < p < 1 \)), we have:
\[
p = \frac{1}{3}
\]

### Final Answer:
The value of \( p \) is \( \frac{1}{3} \).