Verified Solution • 2025–26 Session moderate

QJEEAdv21Ma243f: JEE Advanced 2021 Mathematics - Class 12 JEEAdv21Ma243f Solution: Three numbers are chosen at random, one after another with…

Three numbers are chosen at random, one after another with replacement, from the set S = {1, 2, 3, ......, 100}. Let p1 be the probability that the maximum of chosen numbers is at least 81 and p2 be the probability that the minimum of chosen numbers is at most 40.

The value of 1254p2{{125} \over 4}{p_2} is ___________.

Step-by-Step Verified Solution

p2 = 1 −- p (all three numbers are > 40)

=1−(60100)3=1−27125=98125= 1 - {\left( {{{60} \over {100}}} \right)^3} = 1 - {{27} \over {125}} = {{98} \over {125}}

So, 125p24=1254×98125=984=24.50{{125{p_2}} \over 4} = {{125} \over 4} \times {{98} \over {125}} = {{98} \over 4} = 24.50

Related Learning Resources & Exam Preparation

JEE Advanced PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring