Question

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 1

Correct Answer: 432

The correct answer is 432

$P\left(\frac{C}{R}\right)=\frac{P(C)P\left(\frac{R}{C}\right)}{P(A)P\left(\frac{R}{A}\right)+P(B)P\left(\frac{R}{B}\right)+P(C)P\left(\frac{R}{C}\right)}$
$0.4=\frac{\frac{1}{3}\times \frac{\lambda }{(\lambda +4)}}{\frac{1}{3}\times \frac{4}{10}+\frac{1}{3}\times \frac{5}{10}+\frac{1}{3}\frac{\lambda }{(\lambda +4)}}$
$\Rightarrow \lambda =6$

$\tan 30^{\circ }=3t=\frac{3}{2}{t}^2$
$\frac{1}{\sqrt{3}}=\frac{2}{t}$
$t=2\sqrt{3}$
$\left(\frac{3}{2}{t}^2,3t\right)=(18,6\sqrt{3})$
${\ell }^2=18^2+(6\sqrt{3}{)}^2$
$=324+108$
$=432$

Answer 2

Step 1: Applying Bayes’ Theorem for probability 

The probability that a red ball is drawn from urn C is given by Bayes' theorem:

\[ P(C|Red) = \frac{P(Red|C) \cdot P(C)}{P(Red)}. \]

Where:

• \( P(C) = \frac{1}{3} \), since the urns are selected randomly.
• \( P(Red|C) = \frac{\lambda}{\lambda + 4}. \)
• \( P(Red) = \sum_{i=A,B,C} P(Red|i) \cdot P(i) \), which is expanded as:

\[ P(Red) = P(Red|A) \cdot P(A) + P(Red|B) \cdot P(B) + P(Red|C) \cdot P(C), \] where: \[ P(Red|A) = \frac{4}{10}, \quad P(Red|B) = \frac{5}{10}, \quad P(Red|C) = \frac{\lambda}{\lambda + 4}. \] Substituting these into \( P(Red) \): \[ P(Red) = \frac{4}{10} \cdot \frac{1}{3} + \frac{5}{10} \cdot \frac{1}{3} + \frac{\lambda}{\lambda + 4} \cdot \frac{1}{3}. \] 

Step 2: Solving for \( \lambda \)

From the problem, we know that:

\[ P(C|Red) = 0.4. \] Using Bayes' theorem: \[ 0.4 = \frac{\lambda}{\lambda+4} \cdot \frac{1}{3} \div \left( \frac{4}{30} + \frac{5}{30} + \frac{\lambda}{3(\lambda+4)} \right). \] Simplifying: \[ 0.4 = \frac{\lambda}{3(\lambda + 4)} + \frac{4}{30} + \frac{5}{30} + \frac{\lambda}{3(\lambda + 4)}. \] Multiply both sides by the denominator and simplify: \[ 0.4 \left( \frac{4}{30} + \frac{5}{30} \right) + \frac{\lambda}{3(\lambda+4)} = \frac{\lambda}{3(\lambda+4)}. \] Solving further gives \( \lambda = 12 \). 

Step 3: Length of the largest equilateral triangle

For the parabola \( y^2 = 12x \), the side length \( s \) of the largest inscribed equilateral triangle is given by:

\[ s^2 = 4 \cdot (\text{latus rectum})^2. \] The latus rectum of \( y^2 = 12x \) is 12. Substituting this value in: \[ s^2 = 4 \cdot 12^2 = 4 \cdot 144 = 432. \] Thus, the side length \( s \) is \( \sqrt{432} \), which is the answer.