Three urns A, B and C contain 4 red, 6 black; 5 red, 5 black; and 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

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Solution

Q.
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 __________. [2023]

Ans.
(432)

$P (Red ball from urn C)$

$=$ $\frac{\frac{1}{3} \cdot \frac{λ}{λ + 4}}{\frac{1}{3} \cdot \frac{4}{10} + \frac{1}{3} \cdot \frac{5}{10} + \frac{1}{3} \cdot \frac{λ}{λ + 4}} = 0.4 \Rightarrow \frac{\frac{λ}{λ + 4}}{\frac{4}{10} + \frac{5}{10} + \frac{λ}{λ + 4}} = \frac{4}{10}$

$∴$ $λ = 6$

So, parabola $y^{2} = 6 x$

Let side length of the triangle be $l$ .

$\tan 30 ° = \frac{3 t}{\frac{3}{2} t^{2}}$

$\Rightarrow \frac{1}{\sqrt{3}} = \frac{2}{t}$

$∴$ $t = 2 \sqrt{3}$

So, $(\frac{3}{2} t^{2}, 3 t) = (18, 6 \sqrt{3})$

Now, $l^{2} = 18^{2} + {(6 \sqrt{3})}^{2} = 324 + 108 = 432$

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