The urns A,B and C contains 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…

Let $R$R be the event of drawing red balls.

$$P(\frac{R}{A})=\frac{4}{10}$$P(AR​)=104​

$$P(\frac{R}{B})=\frac{5}{10}$$P(BR​)=105​

$$P(\frac{R}{C})=\frac{\lambda }{4+\lambda }$$P(CR​)=4+λλ​

Now,

$$P(\frac{C}{R})=0.4$$P(RC​)=0.4

$$\Rightarrow \frac{P(C)P(\frac{R}{C})}{P(A)P(\frac{R}{A})+P(B)P(\frac{R}{B})+P(C)P(\frac{R}{C})}=\frac{4}{10}$$⇒P(A)P(AR​)+P(B)P(BR​)+P(C)P(CR​)P(C)P(CR​)​=104​

$$\Rightarrow \frac{\frac{1}{3}\times (\frac{\lambda }{\lambda +4})}{\frac{1}{3}\times \frac{4}{10}+\frac{1}{3}\times \frac{5}{10}+\frac{1}{3}\times (\frac{\lambda }{\lambda +4})}=\frac{4}{10}$$⇒31​×104​+31​×105​+31​×(λ+4λ​)31​×(λ+4λ​)​=104​

$$\Rightarrow \frac{(\frac{\lambda }{\lambda +4})}{\frac{9}{10}+(\frac{\lambda }{\lambda +4})}=\frac{4}{10}$$⇒109​+(λ+4λ​)(λ+4λ​)​=104​

$$\Rightarrow \frac{\lambda }{\lambda +4}=\frac{4}{10}[\frac{9}{10}+(\frac{\lambda }{\lambda +4})]$$⇒λ+4λ​=104​[109​+(λ+4λ​)]

$$\Rightarrow \frac{5\lambda }{\lambda +4}=2[\frac{9}{10}+(\frac{\lambda }{\lambda +4})]$$⇒λ+45λ​=2[109​+(λ+4λ​)]

$$\Rightarrow \frac{5\lambda }{\lambda +4}=\frac{19\lambda +36}{5(\lambda +4)}$$⇒λ+45λ​=5(λ+4)19λ+36​

$$\Rightarrow 25\lambda =19\lambda +36$$⇒25λ=19λ+36

$$\Rightarrow 6\lambda =36$$⇒6λ=36

$$\Rightarrow \lambda =6$$⇒λ=6

So, parabola is

$${y}^{2}=6x=4\times \frac{2}{3}\times x$$y2=6x=4×32​×x

Comparing with ${y}^{2}=4ax$y2=4ax, we get

$$a=\frac{2}{3}$$a=32​

Hence, we have

Let $P\equiv (\frac{2}{3}{t}^{2},\frac{4}{3}t)$P≡(32​t2,34​t)

In $\Delta OPQ$ΔOPQ, we have

$$\mathrm{tan}(30^{\circ})=\frac{2at}{a{t}^{2}}=\frac{2}{t}$$tan(30∘)=at22at​=t2​

$$\Rightarrow \frac{2}{t}=\frac{1}{\sqrt{3}}$$⇒t2​=3​1​

$$\Rightarrow t=2\sqrt{3}$$⇒t=23​

Side length of triangle$=4at=12\sqrt{3}$=4at=123​

Square of side of triangle

$={(12\sqrt{3})}^{2}=432$=(123​)2=432 units