Probability of Integer Area Triangle Formed by Points in Given Region

To find the probability that three distinct points $P$, $Q$, and $R$ chosen from the set $X$ form a triangle with a positive integer area, we first determine the elements of the set $X$.

The set $X$ is defined as: $X = \left\{(x, y) \in \mathbb{Z} \times \mathbb{Z} : \frac{x^2}{8} + \frac{y^2}{20} < 1 \text{ and } y^2 < 5x\right\}$

Step 1: Find the points in $X$

From the condition $y^2 < 5x$, since $y^2 \ge 0$, we must have $x > 0$. Since $x \in \mathbb{Z}$, $x \ge 1$.

From the ellipse inequality $\frac{x^2}{8} + \frac{y^2}{20} < 1$, we observe: $\frac{x^2}{8} < 1 \implies x^2 < 8 \implies x \in \{1, 2\}$

1. For $x = 1$:

   • The condition $y^2 < 5x$ becomes $y^2 < 5 \implies y \in \{-2, -1, 0, 1, 2\}$.
   • Checking the ellipse condition: $\frac{1}{8} + \frac{y^2}{20} < 1 \implies y^2 < \frac{7 \times 20}{8} = 17.5$, which holds true for all $y \in \{-2, -1, 0, 1, 2\}$.
   • So, there are $5$ points on the line $x = 1$: $\{(1, -2), (1, -1), (1, 0), (1, 1), (1, 2)\}$

2. For $x = 2$:

   • The condition $y^2 < 5x$ becomes $y^2 < 10 \implies y \in \{-3, -2, -1, 0, 1, 2, 3\}$.
   • Checking the ellipse condition: $\frac{4}{8} + \frac{y^2}{20} < 1 \implies \frac{y^2}{20} < \frac{1}{2} \implies y^2 < 10$, which holds true for all $y \in \{-3, -2, -1, 0, 1, 2, 3\}$.
   • So, there are $7$ points on the line $x = 2$: $\{(2, -3), (2, -2), (2, -1), (2, 0), (2, 1), (2, 2), (2, 3)\}$

Thus, the total number of points in $X$ is: $n(X) = 5 + 7 = 12$

Step 2: Total number of ways to choose 3 points

The total number of ways to choose $3$ distinct points from $X$ is: $N_{\text{total}} = \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$

Step 3: Find the number of favorable outcomes

Any three points chosen can either have $2$ points on $x = 1$ and $1$ point on $x = 2$, or $1$ point on $x = 1$ and $2$ points on $x = 2$. (Note: Choosing all $3$ points on the same vertical line yields an area of $0$, which is not positive).

The area $\Delta$ of a triangle formed by two points $(x_1, y_1), (x_1, y_2)$ on one line and a third point $(x_2, y_3)$ on the other line is: $\Delta = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} |y_1 - y_2| \times |x_2 - x_1| = \frac{|y_1 - y_2|}{2} \quad (\text{since } |2 - 1| = 1)$

For $\Delta$ to be a positive integer, the difference $|y_1 - y_2|$ must be a positive even integer. This occurs if and only if $y_1$ and $y_2$ have the same parity (both even or both odd).

Case 1: Two points on $x = 1$ and one point on $x = 2$

On the line $x = 1$, $y \in \{-2, -1, 0, 1, 2\}$:

• Even $y$-coordinates: $\{-2, 0, 2\}$ ($3$ points)
• Odd $y$-coordinates: $\{-1, 1\}$ ($2$ points)

Number of ways to choose $2$ points on $x = 1$ with the same parity: $\binom{3}{2} + \binom{2}{2} = 3 + 1 = 4$

For each pair, there are $7$ choices for the third point on $x = 2$: $N_1 = 4 \times 7 = 28$

Case 2: One point on $x = 1$ and two points on $x = 2$

On the line $x = 2$, $y \in \{-3, -2, -1, 0, 1, 2, 3\}$:

• Even $y$-coordinates: $\{-2, 0, 2\}$ ($3$ points)
• Odd $y$-coordinates: $\{-3, -1, 1, 3\}$ ($4$ points)

Number of ways to choose $2$ points on $x = 2$ with the same parity: $\binom{3}{2} + \binom{4}{2} = 3 + 6 = 9$

For each pair, there are $5$ choices for the third point on $x = 1$: $N_2 = 9 \times 5 = 45$

Step 4: Calculate the total favorable outcomes and probability

Total number of favorable combinations: $N_{\text{favorable}} = N_1 + N_2 = 28 + 45 = 73$

Therefore, the required probability $P$ is: $P = \frac{N_{\text{favorable}}}{N_{\text{total}}} = \frac{73}{220}$

Thus, the correct option is B.