JEE MAIN - Mathematics (2023 - 24th January Evening Shift - No. 22)
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
432
Explanation
$E_{1}$ : Ball is drawn from urn $A(4 R+6 B)$
$E_{2}:$ Ball is drawn from urn $B(5 R+5 B)$
$E_{3}$ : Ball is drawn from urn $C(\lambda R+4 B)$
$A \rightarrow$ Ball drawn is red.
Required probability $=P\left(\frac{E_{3}}{A}\right)$
$=\frac{\frac{1}{3} \times \frac{\lambda}{\lambda+4}}{\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}+\frac{1}{3} \times \frac{\lambda}{\lambda+4}}=\frac{2}{5}$
$\Rightarrow\frac{10 \lambda}{19 \lambda+36}=\frac{2}{5}$
$\Rightarrow \lambda=6$
So, parabola $y^2=6 x$
Let side length of the triangle be $l$.
_24th_January_Evening_Shift_en_22_1.png)
$$ \begin{aligned} & \tan 30^{\circ}=\frac{3 t} {\frac{3}{2} t^2} \\\\ & \Rightarrow \frac{1}{\sqrt{3}}=\frac{2}{t} \\\\ & \therefore t=2 \sqrt{3} \\\\ & \text { So, }\left(\frac{3}{2} t^2, 3 t\right) \\\\ & =(18,6 \sqrt{3}) \end{aligned} $$
$$ \text { Now, } l^2=18^2+(6 \sqrt{3})^2=324+108=432 $$
$E_{2}:$ Ball is drawn from urn $B(5 R+5 B)$
$E_{3}$ : Ball is drawn from urn $C(\lambda R+4 B)$
$A \rightarrow$ Ball drawn is red.
Required probability $=P\left(\frac{E_{3}}{A}\right)$
$=\frac{\frac{1}{3} \times \frac{\lambda}{\lambda+4}}{\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}+\frac{1}{3} \times \frac{\lambda}{\lambda+4}}=\frac{2}{5}$
$\Rightarrow\frac{10 \lambda}{19 \lambda+36}=\frac{2}{5}$
$\Rightarrow \lambda=6$
So, parabola $y^2=6 x$
Let side length of the triangle be $l$.
$$ \begin{aligned} & \tan 30^{\circ}=\frac{3 t} {\frac{3}{2} t^2} \\\\ & \Rightarrow \frac{1}{\sqrt{3}}=\frac{2}{t} \\\\ & \therefore t=2 \sqrt{3} \\\\ & \text { So, }\left(\frac{3}{2} t^2, 3 t\right) \\\\ & =(18,6 \sqrt{3}) \end{aligned} $$
$$ \text { Now, } l^2=18^2+(6 \sqrt{3})^2=324+108=432 $$
Comments (0)
