Mathematics - Probability Question with Solution | TestHub

Given,

$P\left(w_n\right)=\frac{P\left(w_{n-1}\right)}{2}, n\ge 2$

Let $P\left(w_1\right)=\lambda$ then

$P\left(w_2\right)=\frac{\lambda }{2}$

$P\left(w_3\right)=\frac{\lambda }{2^2}$

$$\begin{gathered} ⋮ ⋮ ⋮ \\ ⋮ ⋮ ⋮ \end{gathered}$$

$P\left(w_n\right)=\frac{\lambda }{2^{n-1}}$

As

$\sum _{k=1}^{\infty }P\left(w_k\right)=1$

$\Rightarrow \lambda +\frac{\lambda }{2}+\frac{\lambda }{2^2}+\frac{\lambda }{2^3}+....=1$

$\Rightarrow \frac{\lambda }{1-\frac{1}{2}}=1\Rightarrow \lambda =\frac{1}{2}$

So, $P\left(w_n\right)=\frac{1}{2^n}$

$A=\left\{2k+3l; k, l\in N\right\}=\left\{5,7,8,9,10\dots \dots \right\}$

$B=\left\{w_n: n\in A\right\}$

$B=\left\{w_5,w_7,w_8,w_9,w_{10},w_{11},\dots \right\}$

$A=\mathbb{N}-\left\{1, 2, 3, 4, 6\right\}$

$\therefore P\left(B\right)=1-\left[P\left(w_1\right)+P\left(w_2\right)+P\left(w_3\right)+P\left(w_4\right)+P\left(w_6\right)\right]$

$=1-\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{64}\right]$

$=1-\left(\frac{32+16+8+4+1}{64}\right)=\frac{3}{64}$