All five letter words are made using all the letters $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}$ and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $W_n$. Let the probability…

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All five letter words are made using all the letters $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}$ and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $W_n$. Let the probability $\mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)$ of choosing the word $\mathrm{W}_{\mathrm{n}}$ satisfy $\mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)=2 \mathrm{P}\left(\mathrm{W}_{\mathrm{n}-1}\right), \mathrm{n} \gt 1$.
If $\mathrm{P}(\mathrm{CDBEA})=\frac{2^\alpha}{2^\beta-1}, \alpha, \beta \in \mathbb{N}$, then $\alpha+\beta$ is equal to : _______

Aicharya · Accepted Answer

Let $\mathrm{P}\left(\mathrm{W}_1ight)=\mathrm{x}$ $\begin{aligned} & \sum_{i=1}^{120} P\left(W_iight)=1 \ & x+2 x+2^2 x+2^3 x+\ldots+2^{119} x=1 \ & \frac{x\left(2^{120}-1ight)}{(2-1)}=1 \Rightarrow x=\frac{1}{2^{120}-1} \qquad...(i) \end{aligned}$ Rank of CDBEA $\begin{aligned} & 	ext { C D B A E }=1 \ & 	ext { C D B E A }=1\end{aligned}$ $\begin{aligned} & 	ext { So, } \mathrm{P}\left(\mathrm{W}_{64}ight)=2 \mathrm{P}\left(\mathrm{W}_{63}ight)=\ldots=2^{63} \mathrm{P}\left(\mathrm{W}_1ight) \ & =\frac{2^{63}}{2^{120}-1} \ & \alpha+\beta=63+120=183\end{aligned}$

All five letter words are made using all the letters $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},…

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