Three urns A, B and C contain 4 red, 6 black ; 5 red, 5 black ; and λ red, 4 black ballsrespectively. One of the urns is selected at random and a ball is drawn. If the ball drawn isred and the probability that it is drawn from urn C is 0.4, then the square of the length of theside of the largest equilateral triangle, inscribed in the parabola y2=λx with one vertex atthe vertex of the parabola, is_

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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__

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answer is 432.

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