Mathematics · Probability

JEE Main 2025 — 3 April, Morning Shift — Question 40

All five letter words are made using all the letters A, B, C, D, 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 $\mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)=2 \mathrm{P}\left(\mathrm{W}_{\mathrm{n}-1}\right), \mathrm{n}>1$ . If $\mathrm{P}(\mathrm{CDBEA})=\frac{2^{\alpha}}{2^{\beta}-1}, \alpha, \beta \in \mathbb{N}$ , then $\alpha+\beta$ is

equal to: $\_\_\_\_\_$

Answer: 183

Numerical answer — enter this value.

Step-by-step solution

Let $\mathrm{P}\left(\mathrm{W}_{1}\right)=\mathrm{x}$

$\sum_{\mathrm{i}=1}^{120} \mathrm{P}\left(\mathrm{W}_{\mathrm{i}}\right)=1$

$x+2 x+2^{2} x+2^{3} x+\ldots+2^{119} x=1$

\begin{gathered} \frac{\mathrm{x}\left(2^{120}-1\right)}{(2-1)}=1 \Rightarrow \mathrm{x}=\frac{1}{2^{120}-1} …(1) \end{gathered}

Rank of CDBEA

A $\_\_\_\_\_$ $=\mid 4=24$

B $\_\_\_\_\_$ $=\mid 4=24$

C A $\_\_\_\_\_$ $=\lfloor 3=6$

C B $\_\_\_\_\_$ $=\underline{3}=6$

C D A $\_\_\_\_\_$ $=2=2$

C D B A E $=1$

C D B E A $=1$

So, $\mathrm{P}\left(\mathrm{W}_{64}\right)=2 \mathrm{P}\left(\mathrm{W}_{63}\right)=\ldots=2^{63} \mathrm{P}\left(\mathrm{W}_{1}\right)$

$=\frac{2^{63}}{2^{120}-1}$

$\alpha+\beta=63+120=183$

Answer key and solution verified before publishing.

Practise Probability

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 3 April, Morning Shift
Subject: Mathematics
Chapter: Probability
Topic: Conditional Probability and Multiplication Theorem