Let X = x ,   y ∈ ℤ × ℤ   :   x 2 8 + y 2 20 < 1   and   y 2 < 5 x . Three distinct point 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
B. 73 220
Given, Equation of ellipse x 2 8 + y 2 20 = 1 And equation of parabola   y 2 = 5 x Now intersection point of ellipse and parabola will be 2 , 10   &   2 , - 10 Now plotting the region x 2 8 + y 2 20 < 1   and   y 2 < 5 x we get, Now randomly chosen points will be between origin and the intersecting point, So, the integer 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 = C 3 12 = 220 Required number of triangle = 4 × C 1 7 + 9 × C 1 5 = 73 Note: Points are taken such a way that distance between two points are multiple of 2 for example if we take points on line x = 1 we can take following points 1 , 2   &   1 , 0 ,   1 , 0   &   1 , - 2 ,   1 , 1   &   1 , - 1 ,   1 , 2   &   1 , - 2 so there will be 4 pairs from line x = 1 and will chose another point from x = 2 line so triangle will have a positive area. Also note that height of the triangle will always be 1 Hence, probability will be = 73 220
Related: Mathematics — Probability · All PYQ Banks