JEE Main 2023 — Mathematics Probability & Statistics

Given,

$$P({w}_{n})=\frac{P({w}_{n-1})}{2},n\geq 2$$P(wn​)=2P(wn−1​)​,n≥2

Let $P({w}_{1})=\lambda$P(w1​)=λ then

$$P({w}_{2})=\frac{\lambda }{2}$$P(w2​)=2λ​

$$P({w}_{3})=\frac{\lambda }{{2}^{2}}$$P(w3​)=22λ​

$$\vdots \vdots \vdots \vdots \vdots \vdots$$⋮⋮⋮⋮⋮⋮

$$P({w}_{n})=\frac{\lambda }{{2}^{n-1}}$$P(wn​)=2n−1λ​

As

$$\sum _{k=1}^{\infty }P({w}_{k})=1$$k=1∑∞​P(wk​)=1

$$\Rightarrow \lambda +\frac{\lambda }{2}+\frac{\lambda }{{2}^{2}}+\frac{\lambda }{{2}^{3}}+....=1$$⇒λ+2λ​+22λ​+23λ​+....=1

$$\Rightarrow \frac{\lambda }{1-\frac{1}{2}}=1\Rightarrow \lambda =\frac{1}{2}$$⇒1−21​λ​=1⇒λ=21​

So, $P({w}_{n})=\frac{1}{{2}^{n}}$P(wn​)=2n1​

$$A={2k+3l;k,l\in N}={5,7,8,9,10\ldots \ldots }$$A=2k+3l;k,l∈N=5,7,8,9,10……

$$B={{w}_{n}:n\in A}$$B=wn​:n∈A

$$B={{w}_{5},{w}_{7},{w}_{8},{w}_{9},{w}_{10},{w}_{11},\ldots }$$B=w5​,w7​,w8​,w9​,w10​,w11​,…

$$A=\mathbb{N}-{1,2,3,4,6}$$A=N−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−[P(w1​)+P(w2​)+P(w3​)+P(w4​)+P(w6​)]

$$=1-[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{64}]$$=1−[21​+41​+81​+161​+641​]

$$=1-(\frac{32+16+8+4+1}{64})=\frac{3}{64}$$=1−(6432+16+8+4+1​)=643​