JEE Main 2023 (Online) 24th January Evening ShiftQuestion 1
The 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 length of the side of largest equilateral triangle, inscribed in the parabola $y^{2}=\lambda x$ with one vertex at vertex of parabola is
Parabola
Answer$432$
Solution
Concept. Find $\lambda$ by Bayes' theorem, then the equilateral triangle has its sides from $O$ at $\pm30^\circ$ to the axis. Formulas used. (1) $P(C\mid R)=\dfrac{P(R\mid C)}{P(R\mid A)+P(R\mid B)+P(R\mid C)}$ (urns equally likely).(2) Point $(at^2,2at)$: slope from $O$ is $\dfrac2t$.Step 1 - Use (1). Let $p=\dfrac{\lambda}{\lambda+4}$: $\dfrac{p}{0.4+0.5+p}=0.4\Rightarrow0.6p=0.36\Rightarrow p=0.6$. $\lambda=0.6\lambda+2.4\Rightarrow\lambda=6$. Step 2 - Parabola. $y^2=6x$, $a=\frac32$. Other vertices $(at^2,\pm2at)$. Step 3 - Use (2). $\dfrac2t=\tan30^\circ=\dfrac1{\sqrt3}\Rightarrow t=2\sqrt3$. Step 4 - Side. Side $=2\cdot2at=4at=4\cdot\frac32\cdot2\sqrt3=12\sqrt3$. Step 5. Square $=144\cdot3=432$. Answer: 432