JEE Advanced 2023 Paper 1 Q06 Mathematics P&C and Probability Probability Hard

JEE Advanced 2023 Paper 1 · Q06 · Probability

Let $X = \left\{(x,y) \in \mathbb{Z} \times \mathbb{Z} : \dfrac{x^2}{8} + \dfrac{y^2}{20} < 1 \text{ and } y^2 < 5x\right\}$. Three distinct points $P$, $Q$ and $R$ are randomly chosen from $X$. Then the probability that $P$, $Q$ and $R$ form a triangle whose area is a positive integer is

  1. A. $\dfrac{71}{220}$
  2. B. $\dfrac{73}{220}$
  3. C. $\dfrac{79}{220}$
  4. D. $\dfrac{83}{220}$
Reveal answer + step-by-step solution

Correct answer:B

Solution

Solving $\dfrac{x^2}{8} + \dfrac{x}{4} < 1$ alongside $y^2 < 5x$ over integer pairs gives $|X| = 12$ points: at $x=1$, $y \in \{-2,-1,0,1,2\}$ (5 points); at $x=2$, $y \in \{-3,-2,-1,0,1,2,3\}$ (7 points). Total ways $= \binom{12}{3} = 220$. A triangle has integer area when (using shoelace) the base × height product is even; favorable count by case analysis = $73$ (choosing $2$ from one column and $1$ from the other with even-difference, etc.). Probability $= 73/220$.

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