The urns A, B and C contain 4 red,6 black; 5 | vedclass

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(D) Let $R$ be the event of drawing a red ball. Using Bayes' theorem: $P(C|R) = \frac{P(C)P(R|C)}{P(A)P(R|A) + P(B)P(R|B) + P(C)P(R|C)}$ Given $P(A) =

Vedclass · Accepted Answer

$432$

Vedclass · Answer

(D) Let $R$ be the event of drawing a red ball. Using Bayes' theorem: $P(C|R) = \frac{P(C)P(R|C)}{P(A)P(R|A) + P(B)P(R|B) + P(C)P(R|C)}$ Given $P(A) = P(B) = P(C) = \frac{1}{3}$,$P(R|A) = \frac{4}{10}$,$P(R|B) = \frac{5}{10}$,$P(R|C) = \frac{\lambda}{\lambda+4}$ and $P(C|R) = 0.4 = \frac{2}{5}$. $\frac{2}{5} = \frac{\frac{1}{3} \cdot \frac{\lambda}{\lambda+4}}{\frac{1}{3}(\frac{4}{10} + \frac{5}{10} + \frac{\lambda}{\lambda+4})} = \frac{\frac{\lambda}{\lambda+4}}{0.9 + \frac{\lambda}{\lambda+4}}$ $0.36 + 0.4 \frac{\lambda}{\lambda+4} = \frac{\lambda}{\lambda+4} \Rightarrow 0.36 = 0.6 \frac{\lambda}{\lambda+4} \Rightarrow \frac{\lambda}{\lambda+4} = 0.6 = \frac{3}{5}$ $5\lambda = 3\lambda + 12 \Rightarrow 2\lambda = 12 \Rightarrow \lambda = 6$. The parabola is $y^2 = 6x$. Let the vertices of the equilateral triangle be $(0,0)$,$(at_1^2, 2at_1)$,and $(at_2^2, 2at_2)$ where $4a = 6 \Rightarrow a = 1.5$. Due to symmetry about the $x$-axis,$t_1 = t$ and $t_2 = -t$. The side length $\ell$ satisfies $\ell^2 = (at^2)^2 + (2at)^2 = a^2t^4 + 4a^2t^2$. Also,the slope of the side is $	an(30^{\circ}) = \frac{2at}{at^2} = \frac{2}{t} = \frac{1}{\sqrt{3}} \Rightarrow t = 2\sqrt{3}$. $\ell^2 = (1.5)^2(2\sqrt{3})^4 + 4(1.5)^2(2\sqrt{3})^2 = 2.25(144) + 9(12) = 324 + 108 = 432$.

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