Verified Solution • 2025–26 Session moderate

QJEEM23Ma8dcf: JEE Main 2023 Mathematics - Class 12 JEEM23Ma8dcf Solution: Three urns A, B and C contain 4 red, 6 black; 5 red, 5 black;…

Three urns A, B and C contain 4 red, 6 black; 5 red, 5 black; and λ\lambda red, 4 black balls respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn is red and the probability that it is drawn from urn C is 0.4 then the square of the length of the side of the largest equilateral triangle, inscribed in the parabola y2=λxy^2=\lambda x with one vertex at the vertex of the parabola, is :

Step-by-Step Verified Solution

E1E_{1} : Ball is drawn from urn A(4R+6B)A(4 R+6 B)

E2:E_{2}: Ball is drawn from urn B(5R+5B)B(5 R+5 B)

E3E_{3} : Ball is drawn from urn C(λR+4B)C(\lambda R+4 B)

A→A \rightarrow Ball drawn is red.

Required probability =P(E3A)=P\left(\frac{E_{3}}{A}\right)

=13×λλ+413×410+13×510+13×λλ+4=25=\frac{\frac{1}{3} \times \frac{\lambda}{\lambda+4}}{\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}+\frac{1}{3} \times \frac{\lambda}{\lambda+4}}=\frac{2}{5}

⇒10λ19λ+36=25\Rightarrow\frac{10 \lambda}{19 \lambda+36}=\frac{2}{5}

⇒λ=6\Rightarrow \lambda=6

So, parabola y2=6xy^2=6 x

Let side length of the triangle be ll .

JEE Main 2023 (Online) 24th January Evening Shift Mathematics - Probability Question 54 English Explanation
amp;tan⁡30∘=3t32t2amp;⇒13=2tamp;∴t=23amp; So, (32t2,3t)amp;=(18,63)\begin{aligned}& \tan 30^{\circ}=\frac{3 t} {\frac{3}{2} t^2} \\\\& \Rightarrow \frac{1}{\sqrt{3}}=\frac{2}{t} \\\\& \therefore t=2 \sqrt{3} \\\\& \text { So, }\left(\frac{3}{2} t^2, 3 t\right) \\\\& =(18,6 \sqrt{3})\end{aligned}

 Now, l2=182+(63)2=324+108=432\text { Now, } l^2=18^2+(6 \sqrt{3})^2=324+108=432

Related Learning Resources & Exam Preparation

JEE Main 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