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(B) Let $P(w_1) = \lambda$. Then $P(w_2) = \frac{\lambda}{2}, P(w_3) = \frac{\lambda}{4}, \ldots, P(w_n) = \frac{\lambda}{2^{n-1}}$.Since $\sum_{k=1}^

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$\frac{3}{64}$

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(B) Let $P(w_1) = \lambda$. Then $P(w_2) = \frac{\lambda}{2}, P(w_3) = \frac{\lambda}{4}, \ldots, P(w_n) = \frac{\lambda}{2^{n-1}}$.Since $\sum_{k=1}^{\infty} P(w_k) = 1$,we have $\sum_{k=1}^{\infty} \frac{\lambda}{2^{k-1}} = 1$.Using the sum of an infinite geometric series,$\frac{\lambda}{1 - 1/2} = 1 \Rightarrow 2\lambda = 1 \Rightarrow \lambda = 1/2$.Thus,$P(w_n) = \frac{1}{2^n}$.The set $A = \{2k + 3\ell : k, \ell \in \mathbb{N}\}$. Since $k, \ell \geq 1$,the smallest values are:For $k=1, \ell=1, n=5$.For $k=2, \ell=1, n=7$.For $k=1, \ell=2, n=8$.For $k=3, \ell=1, n=9$.For $k=2, \ell=2, n=10$.It can be shown that $A = \mathbb{N} \setminus \{1, 2, 3, 4, 6\}$.Therefore,$P(B) = 1 - [P(w_1) + P(w_2) + P(w_3) + P(w_4) + P(w_6)]$.$P(B) = 1 - [\frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \frac{1}{2^6}] = 1 - [\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{64}]$.$P(B) = 1 - [\frac{32 + 16 + 8 + 4 + 1}{64}] = 1 - \frac{61}{64} = \frac{3}{64}$.

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

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