Question

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

• A. $\frac{71}{220}$
• B. $\frac{73}{220}$
• C. $\frac{79}{220}$
• D. $\frac{83}{220}$

Answer 1

Answer: (B)

$\frac{73}{220}$

Answer 2

The points inside region are {(2, 1), (2, –1), (2, 2), (2, –2), (2, 3), (2, –3), (2, 0), (1, 1), (1, –1), (1, 2), (1, –2), (1, 0)}.
Total number of ways to select three points = $^{12}C_3$ = 220
Required number of triangle = 4 × $^7C_1$ + 9 × $^5C_1$ = 73
Points are taken such a way that distance between two points are multiple of 2.
So, the correct option is (B) : $\frac{73}{220}$