Three urns A, B and C contain 4 red, 6 black; 5 red, 5 black; and 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 with one vertex at the vertex of the parabola, is
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Answer: 432
Solution
First, find λ. P(red from C) = 0.4, and by Bayes' theorem:
P(C|red) = P(red|C)·(1/3) / [P(red|A)·(1/3) + P(red|B)·(1/3) + P(red|C)·(1/3)]
= [λ/(λ+4)] / [4/10 + 5/10 + λ/(λ+4)] = 0.4
So: [λ/(λ+4)] / [9/10 + λ/(λ+4)] = 2/5
5λ/(λ+4) = 2[9/10 + λ/(λ+4)]
5λ/(λ+4) = 9/5 + 2λ/(λ+4)
3λ/(λ+4) = 9/5
15λ = 9(λ+4)
6λ = 36 → λ = 6
Now find the largest equilateral triangle inscribed in y² = 6x with one vertex at the origin (vertex of parabola).
By symmetry, the triangle has one vertex at origin and the other two symmetric about the x-axis. A point on the parabola: (at², 2at) where a = 3/2, so points are (3t²/2, 3t) and (3t²/2, -3t).
The side from origin to (3t²/2, 3t) has length L = √[(3t²/2)² + (3t)²] = 3t·√(t
P(C|red) = P(red|C)·(1/3) / [P(red|A)·(1/3) + P(red|B)·(1/3) + P(red|C)·(1/3)]
= [λ/(λ+4)] / [4/10 + 5/10 + λ/(λ+4)] = 0.4
So: [λ/(λ+4)] / [9/10 + λ/(λ+4)] = 2/5
5λ/(λ+4) = 2[9/10 + λ/(λ+4)]
5λ/(λ+4) = 9/5 + 2λ/(λ+4)
3λ/(λ+4) = 9/5
15λ = 9(λ+4)
6λ = 36 → λ = 6
Now find the largest equilateral triangle inscribed in y² = 6x with one vertex at the origin (vertex of parabola).
By symmetry, the triangle has one vertex at origin and the other two symmetric about the x-axis. A point on the parabola: (at², 2at) where a = 3/2, so points are (3t²/2, 3t) and (3t²/2, -3t).
The side from origin to (3t²/2, 3t) has length L = √[(3t²/2)² + (3t)²] = 3t·√(t
[JEE Main 2023 · 24 Jan · Shift 2]