1. Key Concepts and Formulas

• Bayes' Theorem: This theorem helps us calculate conditional probabilities. For events A and B, the probability of A occurring given B has occurred is $P(A|B) = \frac{P(B|A)P(A)}{P(B)}$.
• Total Probability Theorem: If $A_1, A_2, \dots, A_n$ are mutually exclusive and exhaustive events, then the probability of an event B is $P(B) = \sum_{i=1}^n P(B|A_i)P(A_i)$.
• Parabola Properties: The standard form of a parabola opening to the right with its vertex at the origin is $y^2 = 4ax$. For $y^2 = \lambda x$, the vertex is at $(0,0)$, and the axis of symmetry is the x-axis.
• Equilateral Triangle Properties: All sides are equal in length, and all interior angles are $60^\circ$. For an equilateral triangle inscribed in a parabola $y^2=kx$ (which is symmetric about the x-axis) with one vertex at the origin, the "largest" such triangle will have its other two vertices symmetric with respect to the x-axis. The lines connecting the origin to these two vertices will therefore make angles of $\pm 30^\circ$ with the x-axis.

2. Step-by-Step Solution

This problem has two main parts: first, determining the value of $\lambda$ using probability concepts, and second, using this value of $\lambda$ to find the square of the side length of an equilateral triangle inscribed in a parabola.

Part 1: Determining the Value of $\lambda$ (Probability)

Let's define the events for clarity:

• $A$: Urn A is selected.
• $B$: Urn B is selected.
• $C$: Urn C is selected.
• $R$: A red ball is drawn.

Step 1: State the initial probabilities of selecting each urn. Since one of the three urns is selected at random, the probability of selecting each urn is equal: $P(A) = P(B) = P(C) = \frac{1}{3}$

Step 2: Determine the conditional probabilities of drawing a red ball from each urn.

• Urn A contains 4 red and 6 black balls, totaling 10 balls. $P(R|A) = \frac{\text{Number of red balls in A}}{\text{Total balls in A}} = \frac{4}{10}$
• Urn B contains 5 red and 5 black balls, totaling 10 balls. $P(R|B) = \frac{\text{Number of red balls in B}}{\text{Total balls in B}} = \frac{5}{10}$
• Urn C contains $\lambda$ red and 4 black balls, totaling $\lambda+4$ balls. $P(R|C) = \frac{\text{Number of red balls in C}}{\text{Total balls in C}} = \frac{\lambda}{\lambda+4}$

Step 3: Calculate the total probability of drawing a red ball, $P(R)$. We use the Total Probability Theorem, as drawing a red ball can occur by selecting any of the three urns: $P(R) = P(R|A)P(A) + P(R|B)P(B) + P(R|C)P(C)$ Substitute the probabilities from Step 1 and Step 2: $P(R) = \left(\frac{4}{10}\right)\left(\frac{1}{3}\right) + \left(\frac{5}{10}\right)\left(\frac{1}{3}\right) + \left(\frac{\lambda}{\lambda+4}\right)\left(\frac{1}{3}\right)$ Factor out the common term $\frac{1}{3}$: $P(R) = \frac{1}{3} \left( \frac{4}{10} + \frac{5}{10} + \frac{\lambda}{\lambda+4} \right)$ $P(R) = \frac{1}{3} \left( \frac{9}{10} + \frac{\lambda}{\lambda+4} \right)$

Step 4: Apply Bayes' Theorem using the given information to find $\lambda$. We are given that the probability that the ball drawn is from urn C, given that it is red, is $P(C|R) = 0.4$. Using Bayes' Theorem: $P(C|R) = \frac{P(R|C)P(C)}{P(R)}$ Substitute the expressions we found for $P(R|C)$, $P(C)$, and $P(R)$: $0.4 = \frac{\left(\frac{\lambda}{\lambda+4}\right)\left(\frac{1}{3}\right)}{\frac{1}{3} \left( \frac{9}{10} + \frac{\lambda}{\lambda+4} \right)}$ The term $\frac{1}{3}$ cancels out from the numerator and denominator: $0.4 = \frac{\frac{\lambda}{\lambda+4}}{\frac{9}{10} + \frac{\lambda}{\lambda+4}}$ To simplify, let $X = \frac{\lambda}{\lambda+4}$. Also, express $0.4$ as a fraction: $0.4 = \frac{4}{10} = \frac{2}{5}$. $\frac{2}{5} = \frac{X}{\frac{9}{10} + X}$ Multiply both sides by $\left(\frac{9}{10} + X\right)$: $2\left(\frac{9}{10} + X\right) = 5X$ $\frac{18}{10} + 2X = 5X$ $\frac{9}{5} = 3X$ Divide by 3 to solve for $X$: $X = \frac{9}{15} = \frac{3}{5}$

Step 5: Solve for $\lambda$. Now, substitute back $X = \frac{\lambda}{\lambda+4}$: $\frac{\lambda}{\lambda+4} = \frac{3}{5}$ Cross-multiply to solve for $\lambda$: $5\lambda = 3(\lambda+4)$ $5\lambda = 3\lambda + 12$ $2\lambda = 12$ $\lambda = 6$ Since $\lambda$ represents the number of red balls, it must be a non-negative integer. Our value $\lambda=6$ is consistent with this requirement.

Part 2: Finding the Square of the Side Length of the Equilateral Triangle (Coordinate Geometry)

Now that we have $\lambda=6$, the equation of the parabola is $y^2 = 6x$.

Step 1: Identify the vertices of the equilateral triangle. The problem states that one vertex of the equilateral triangle is at the vertex of the parabola. For $y^2 = 6x$, the vertex is at the origin $O(0,0)$. Let the other two vertices be $P$ and $Q$. For the "largest" equilateral triangle inscribed in $y^2=6x$ with one vertex at the origin, the other two vertices must be symmetric with respect to the x-axis (the axis of symmetry of the parabola). So, if $P$ has coordinates $(x,y)$, then $Q$ must have coordinates $(x,-y)$.

Step 2: Use properties of an equilateral triangle to relate $x$ and $y$. For an equilateral triangle $OPQ$ with $O(0,0)$, $P(x,y)$, and $Q(x,-y)$, the side lengths $OP$, $OQ$, and $PQ$ must all be equal. The length of side $OP$ is $\sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2+y^2}$. The length of side $PQ$ is $\sqrt{(x-x)^2 + (y-(-y))^2} = \sqrt{0^2 + (2y)^2} = \sqrt{4y^2} = 2|y|$. Equating the squares of these lengths: $x^2 + y^2 = (2y)^2$ $x^2 + y^2 = 4y^2$ $x^2 = 3y^2$ This relationship implies that $y = \pm \frac{x}{\sqrt{3}}$. This means the lines $OP$ and $OQ$ make angles of $\pm \arctan\left(\frac{1}{\sqrt{3}}\right) = \pm 30^\circ$ with the x-axis, confirming our initial geometric insight for the "largest" triangle.

Step 3: Substitute the relationship into the parabola equation to find $x$. The point $P(x,y)$ lies on the parabola $y^2 = 6x$. From Step 2, we have $y^2 = \frac{x^2}{3}$. Substitute this into the parabola equation: $\frac{x^2}{3} = 6x$ Multiply by 3: $x^2 = 18x$ Rearrange the equation: $x^2 - 18x = 0$ Factor out $x$: $x(x - 18) = 0$ This gives two possible values for $x$: $x=0$ or $x=18$. Since $P$ is a distinct vertex from the origin $O(0,0)$, its x-coordinate cannot be $0$. Therefore, $x = 18$.

Step 4: Find the y-coordinate of vertices $P$ and $Q$. Using the relation $y^2 = \frac{x^2}{3}$ and $x=18$: $y^2 = \frac{18^2}{3} = \frac{324}{3} = 108$ $y = \pm\sqrt{108} = \pm\sqrt{36 \times 3} = \pm 6\sqrt{3}$ So the vertices are $O(0,0)$, $P(18, 6\sqrt{3})$, and $Q(18, -6\sqrt{3})$.

Step 5: Calculate the square of the side length ($s^2$). The side length $s$ is the distance $OP$. $s^2 = x^2 + y^2$ Substitute $x=18$ and $y=6\sqrt{3}$: $s^2 = (18)^2 + (6\sqrt{3})^2$ $s^2 = 324 + (36 \times 3)$ $s^2 = 324 + 108$ $s^2 = 432$

3. Common Mistakes & Tips

• Validating $\lambda$: Always ensure that the value of $\lambda$ derived from the probability section makes sense in the context of the problem (e.g., it must be a non-negative integer when representing the number of balls).
• Non-zero vertex: When solving for $x$ in $x(x-18)=0$, remember that $x=0$ corresponds to the origin (the vertex already identified), so the other distinct vertices must have $x \neq 0$.
• Geometric interpretation: For problems involving inscribed shapes, visualizing the symmetry and angular properties (like the $30^\circ$ angle for the equilateral triangle here) can significantly simplify calculations.

4. Summary

We began by applying Bayes' Theorem and the Total Probability Theorem to the given information about drawing a red ball from urn C. This allowed us to determine the value of $\lambda$, the number of red balls in urn C, which we found to be 6. Next, we used this value of $\lambda$ to define the parabola as $y^2=6x$. Considering an equilateral triangle with one vertex at the origin (the parabola's vertex) and the other two vertices symmetrically placed on the parabola about the x-axis, we derived the coordinates of these vertices. Using the properties of an equilateral triangle, we established a relationship between the x and y coordinates ($x^2=3y^2$). Substituting this into the parabola's equation, we found the coordinates of the other vertices to be $(18, \pm 6\sqrt{3})$. Finally, we calculated the square of the side length ($s^2$) of this triangle using the distance formula, which resulted in $s^2 = 432$.

5. Final Answer

The final answer is $\boxed{1}$.