The urns A, B, and C contain 4 red, 6 black; 5 red, 5 black; and red, 4 black balls respec…

Question

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 was 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

Accepted Answer

By Bayes' theorem, with $t=\dfrac{\lambda}{\lambda+4}$:
$$P(C\mid R)=\frac{t}{0.4+0.5+t}=0.4\ \Longrightarrow\ 0.6t=0.36\ \Longrightarrow\ t=0.6\ \Longrightarrow\ \lambda=6.$$

For the largest equilateral triangle inscribed in $y^2=6x$ with one vertex at the origin, by symmetry the other two vertices are $(x,\pm y)$ with $y/x=\tan30^\circ=1/\sqrt3$. Combined with $y^2=6x$: $\dfrac{x^2}{3}=6x\Rightarrow x=18,\ y=6\sqrt3$.

$$(\text{side})^2=x^2+y^2=324+108=432.$$

Solutions · 1