Let $S=\{w_1,w_2,...\}$ . 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 $P(B)=?$

Question

Let $S=\{w_1,w_2,...\}$ 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 $P(B)=$

(A)$\frac{3}{32}$ (B)$\frac{1}{16}$ (C)$\frac{3}{64}$ (D)$\frac{1}{32}$

My Attempt: I am quite confused here as to what $P(w_1)$ is.

Set $A$ contains all natural numbers greater than $4$ (excluding $6$)and

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

$P(B)=1-\left(P(w_2)+P(w_3)+P(w_4)+P(w_6)\right)=1-(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{32})=\frac{3}{32}$

which is the option (A).

But why should $P(w_1)$ be $1$.

Also I am not sure whether (A) is the correct option.