Let $$S = \{w_1, w_2, \ldots\}$$ be the sample space associated to a random experiment. Let $$P(w_n) = \frac{P(w_{n-1})}{2}$$, $$n \geq 2$$. Let $$A = \{2k + 3l; k, l \in \mathbb{N}\}$$ and $$B = \{w_n; n \in A\}$$. Then $$P(B)$$ is equal to

Step 1: Determine the probabilities of the individual sample points

We are given that $$S = \{w_1, w_2, w_3, \ldots\}$$ is the sample space of a random experiment. The relation between successive probabilities is:

$$P(w_n) = \frac{P(w_{n-1})}{2} \quad \text{for} \quad n \geq 2$$

This forms a geometric sequence with a common ratio of $$\frac{1}{2}$$:

$$P(w_1) = P(w_1)$$

$$P(w_2) = \frac{P(w_1)}{2}$$

$$P(w_3) = \frac{P(w_1)}{2^2}$$

$$P(w_n) = \frac{P(w_1)}{2^{n-1}}$$

Since the sum of all probabilities in a sample space must equal 1:

$$\sum_{n=1}^{\infty} P(w_n) = 1$$

$$P(w_1) \left(1 + \frac{1}{2} + \frac{1}{2^2} + \dots\right) = 1$$

Using the sum of an infinite geometric progression formula $$S_{\infty} = \frac{a}{1-r}$$:

$$P(w_1) \cdot \left(\frac{1}{1 - \frac{1}{2}}\right) = 1$$

$$2 P(w_1) = 1 \implies P(w_1) = \frac{1}{2}$$

Substituting $$P(w_1)$$ back into the general term gives:

$$P(w_n) = \frac{1}{2^n}$$

Step 2: Identify the elements of set A

The set $$A$$ is defined for natural numbers $$k, l \in \mathbb{N}$$ ($$k, l \geq 1$$) as:

$$A = \{2k + 3l; \, k, l \in \mathbb{N}\}$$

Let us find the possible values generated by this linear combination starting from the minimum values:

- For $$k = 1, l = 1 \implies 2(1) + 3(1) = 5$$

- For $$k = 2, l = 1 \implies 2(2) + 3(1) = 7$$

- For $$k = 1, l = 2 \implies 2(1) + 3(2) = 8$$

- For $$k = 3, l = 1 \implies 2(3) + 3(1) = 9$$

- For $$k = 2, l = 2 \implies 2(2) + 3(2) = 10$$

By analyzing the values, we observe that:

- The numbers $$1, 2, 3, 4$$ cannot be formed because the minimum value is 5.

- The number $$6$$ cannot be formed because if $$2k + 3l = 6$$, since $$l \geq 1$$, the only options are $$l=1 \implies 2k=3$$ (no integer solution) or $$l \geq 2 \implies 2k \leq 0$$ (not a natural number).

- All other natural numbers greater than or equal to 5 can be formed.

Thus, the set $$A$$ contains all natural numbers except $$\{1, 2, 3, 4, 6\}$$:

$$A = \{5, 7, 8, 9, 10, 11, \ldots\}$$

Step 3: Calculate the probability of event B

The event $$B$$ is defined as $$B = \{w_n; \, n \in A\}$$. The total probability $$P(B)$$ is the sum of probabilities of $$w_n$$ for all $$n \in A$$:

$$P(B) = \sum_{n \in A} P(w_n) = \sum_{n \in A} \frac{1}{2^n}$$

We can compute this by taking the sum of the entire infinite series from $$n = 1$$ to $$\infty$$ and subtracting the probabilities of the missing indices $$\{1, 2, 3, 4, 6\}$$:

$$P(B) = \sum_{n=1}^{\infty} \frac{1}{2^n} - \left(\frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \frac{1}{2^6}\right)$$

We know that the full sum evaluates to:

$$\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1$$

Now, calculate the sum of the excluded terms:

$$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{64} = \frac{32 + 16 + 8 + 4 + 1}{64} = \frac{61}{64}$$

Subtracting this from 1:

$$P(B) = 1 - \frac{61}{64} = \frac{3}{64}$$

Conclusion:

The value of $$P(B)$$ is equal to $$\frac{3}{64}$$.