JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Probability

With equal probability $\frac{1}{3}$ of selecting each urn, Bayes' theorem gives $P(C\mid R)=\frac{\frac{\lambda }{\lambda +4}}{\frac{4}{10}+\frac{5}{10}+\frac{\lambda }{\lambda +4}}=0.4$. Writing $u=\frac{\lambda }{\lambda +4}$, this is $0.4=\frac{u}{0.9+u}$, so $u=0.6$ and $\lambda =6$. The parabola is then $y^2=6x$. An equilateral triangle with one vertex at the origin and the opposite side perpendicular to the axis has its other vertices at $\left(\frac{y_0^2}{6},\pm y_0\right)$; the $30^{\circ }$ half-angle at the origin gives $\tan 30^{\circ }=\frac{y_0}{y_0^2/6}=\frac{6}{y_0}$, so $y_0=6\sqrt{3}$ and $\frac{y_0^2}{6}=18$. The side is the distance from the origin to $(18,6\sqrt{3})$, whose square is $18^2+(6\sqrt{3}{)}^2=324+108=432$. The square of the side length is $432$.