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(B) Given the set $X = \{(x, y) \in \mathbb{Z} 	imes \mathbb{Z} : \frac{x^2}{8} + \frac{y^2}{20} < 1 	ext{ and } y^2 < 5x\}$.Solving the boundary eq

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$\frac{73}{220}$

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(B) Given the set $X = \{(x, y) \in \mathbb{Z} 	imes \mathbb{Z} : \frac{x^2}{8} + \frac{y^2}{20} Solving the boundary equations $\frac{x^2}{8} + \frac{y^2}{20} = 1$ and $y^2 = 5x$,we substitute $y^2 = 5x$ into the ellipse equation: $\frac{x^2}{8} + \frac{5x}{20} = 1 \implies \frac{x^2}{8} + \frac{x}{4} = 1 \implies x^2 + 2x - 8 = 0 \implies (x+4)(x-2) = 0$. Since $y^2  0$,so $x = 2$. For $x = 2$,$y^2 The set $X$ contains $12$ points: $X = \{(1, -2), (1, -1), (1, 0), (1, 1), (1, 2), (2, -3), (2, -2), (2, -1), (2, 0), (2, 1), (2, 2), (2, 3)\}$.The total number of ways to choose $3$ points is $n(S) = {}^{12}C_3 = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$.For a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$,the area is $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. Since points lie on lines $x=1$ and $x=2$,the base is either on $x=1$ or $x=2$ (length $|y_i - y_j|$) or the points are distributed across both lines. The area is an integer if the base length is even.Triangles with base on $x=1$ (length $d$): $d=2$ ($4$ pairs),$d=4$ ($2$ pairs). Triangles with base on $x=2$ (length $d$): $d=2$ ($6$ pairs),$d=4$ ($4$ pairs),$d=6$ ($2$ pairs).Calculating favorable cases: $46 + 22 + 5 = 73$.Thus,the probability is $\frac{73}{220}$.

Let $X = \{(x, y) \in \mathbb{Z} \times \mathbb{Z} : \frac{x^2}{8} + \frac{y^2}{20} < 1 \text{ and } y^2 < 5x\}$. 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

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