Three urns A, B and C contain 4 red, 6 black ; 5 red, 5 black ; and $λ$ 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 $y^{2} = λ x$ with one vertex at
the vertex of the parabola, is__

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 432.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Urn	Red	Black
A	4	6
B	5	5
C	$λ$	4

Let E be the event to set red ball from ‘C’
$P (C / E) = \frac{P (C) \cdot P (E / C)}{P (A) P (E / A) + P (B) P (E / B) + P (C) P (E / C)}$
Given $0.4 = \frac{\frac{1}{3} \times \frac{λ}{λ + 4}}{\frac{1}{3} (\frac{4}{10} + \frac{5}{10} + \frac{λ}{λ + 4})}$
$\begin{array}{l} \frac{4}{10} (\frac{9}{10} + \frac{λ}{λ + 4}) = \frac{λ}{λ + 4} \\ 76 λ + 144 = 100 λ \Rightarrow 24 λ = 144 \Rightarrow λ = 6 \end{array}$
Given parabola $y^{2} = λ x \Rightarrow y^{2} = 6 x$
Question Image

$\begin{array}{l} \sin 30^{\circ} = \frac{BD}{a} \Rightarrow BD = \frac{a}{2} \\ \cos 30^{\circ} = \frac{AD}{a} \Rightarrow AD = \frac{\sqrt{3}}{2} a \\ B = (\frac{\sqrt{3}}{2} a, \frac{a}{2}) lies on parabola y^{2} = 6 x \\ \frac{a^{2}}{4} = 6 (\frac{\sqrt{3} a}{2}) \Rightarrow a = 12 \sqrt{3} \\ a^{2} = 144 \times 3 \Rightarrow a^{2} = 432 \end{array}$