JEE Mains · Maths · STD 11 - 14. probability
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 : _______
- A 180
- B 181
- C 182
- D 183
Answer & Solution
Correct Answer
(D) 183
Step-by-step Solution
Detailed explanation
Let \(\mathrm{P}\left(\mathrm{W}_1\right)=\mathrm{x}\) \(\begin{aligned} & \sum_{i=1}^{120} P\left(W_i\right)=1 \\ & x+2 x+2^2 x+2^3 x+\ldots+2^{119} x=1 \\ & \frac{x\left(2^{120}-1\right)}{(2-1)}=1 \Rightarrow x=\frac{1}{2^{120}-1} \qquad...(i) \end{aligned}\) Rank of CDBEA…
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