Let S=w_1,w_2, be a sample space with P(w_n)=P(w_n-1)2, n2.

Question

Let $S=\{w_1,w_2,\dots\}$ be a sample space with $P(w_n)=\dfrac{P(w_{n-1})}{2}$, $n\ge2$. Let $A=\{2k+3\ell: k,\ell\in\mathbb N\}$ and $B=\{w_n: n\in A\}$. Then $P(B)$ is equal to

Accepted Answer

$P(w_1)=\lambda$, $P(w_n)=\dfrac{\lambda}{2^{n-1}}$. Since $\sum P(w_n)=1$, $\lambda=\dfrac12$, so $P(w_n)=\dfrac{1}{2^n}$.

$A=\{2k+3\ell\}=\mathbb N\setminus\{1,2,3,4,6\}$ (5 excluded values).

$$P(B) = 1-\left[\frac12+\frac14+\frac18+\frac{1}{16}+\frac{1}{64}\right] = 1-\frac{32+16+8+4+1}{64}=\frac{3}{64}$$

Solutions · 1