Mathematics · 3D Geometry

JEE Advanced 2021 — Paper 1 — Question 44

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

The value of $\frac{125}{4}{{p}_{2}}$ is $\text{ }\!\!~\!\!\text{ }$

Answer: 24.5

Numerical answer — enter this value.

Step-by-step solution

${{\text{P}}_{2}}=1$ - (Probability that 3 chosen numbers are greater than 40) $=1-{{\left( \frac{60}{100} \right)}^{3}}=1-{{\left( \frac{3}{5} \right)}^{3}}=\frac{98}{125}$

So, $\frac{125}{4}$ ${{P}_{2}}=\frac{125}{4}\times \frac{98}{125}=24.5$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2021
Paper: Paper 1
Subject: Mathematics
Chapter: 3D Geometry
Topic: Family of planes, bisector planes.