Question

Three urns $A,B$ and $C$ contain $4$ red, $6$ black; $5$ red, $5$ black; and $\lambda$ 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 $y^2=\lambda x$ with one vertex at the vertex of the parabola, is

Answer 1

Answer: 432

$P\left(\frac{C}{R}\right)=\frac{P(C)P\left(\frac{R}{C}\right)}{P(A)P\left(\frac{R}{A}\right)+P(B)P\left(\frac{R}{B}\right)+P(C)P\left(\frac{R}{C}\right)}$
$0.4=\frac{\frac{1}{3}\times \frac{\lambda }{(\lambda +4)}}{\frac{1}{3}\times \frac{4}{10}+\frac{1}{3}\times \frac{5}{10}+\frac{1}{3}\frac{\lambda }{(\lambda +4)}}$
$\Rightarrow \lambda =6$

$\tan 30^{\circ }=3t=\frac{3}{2}{t}^2$
$\frac{1}{\sqrt{3}}=\frac{2}{t}$
$t=2\sqrt{3}$
$\left(\frac{3}{2}{t}^2,3t\right)=(18,6\sqrt{3})$
${\ell }^2=18^2+(6\sqrt{3}{)}^2$
$=324+108$
$=432$