JEE Main 2025 — Probability Question with Solution
JEE Main 2025 (3 Apr Shift 1)
Question
All five letter words are made using all the letters and arranged as in an English dictionary with serial numbers. Let the word at serial number be denoted by . Let the probability of choosing the word satisfy .
If , then is equal to : _______
If , then is equal to : _______
Enter your answer
Show full solutionCorrect answer: 183
Correct answer
183
Step-by-step explanation
Let
$\begin{aligned}
& \sum_{i=1}^{120} P\left(W_i\right)=1 \\ & x+2 x+2^2 x+2^3 x+\ldots+2^{119} x=1 \\ & \frac{x\left(2^{120}-1\right)}{(2-1)}=1 \Rightarrow x=\frac{1}{2^{120}-1} \qquad...(i)
\end{aligned}\begin{aligned} & \text { C D B A E }=1 \\ & \text { C D B E A }=1\end{aligned}\begin{aligned} & \text { So, } \mathrm{P}\left(\mathrm{W}_{64}\right)=2 \mathrm{P}\left(\mathrm{W}_{63}\right)=\ldots=2^{63} \mathrm{P}\left(\mathrm{W}_1\right) \\ & =\frac{2^{63}}{2^{120}-1} \\ & \alpha+\beta=63+120=183\end{aligned}$
$\begin{aligned}
& \sum_{i=1}^{120} P\left(W_i\right)=1 \\ & x+2 x+2^2 x+2^3 x+\ldots+2^{119} x=1 \\ & \frac{x\left(2^{120}-1\right)}{(2-1)}=1 \Rightarrow x=\frac{1}{2^{120}-1} \qquad...(i)
\end{aligned}\begin{aligned} & \text { C D B A E }=1 \\ & \text { C D B E A }=1\end{aligned}\begin{aligned} & \text { So, } \mathrm{P}\left(\mathrm{W}_{64}\right)=2 \mathrm{P}\left(\mathrm{W}_{63}\right)=\ldots=2^{63} \mathrm{P}\left(\mathrm{W}_1\right) \\ & =\frac{2^{63}}{2^{120}-1} \\ & \alpha+\beta=63+120=183\end{aligned}$
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This is a previous-year question from JEE Main 2025, covering the Probability chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.