JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Probability

Let

X = {(x, y) \in Z \times Z : \frac{x ^{2}}{8} + \frac{y ^{2}}{20} < 1

and

y^{2} < 5 x}

. 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) $\frac{73}{220}$ ✓ Correct
(C) $\frac{79}{220}$
(D) $\frac{83}{220}$

Answer: (B)

Solution

The region X is defined by

\frac{x ^{2}}{8} + \frac{y ^{2}}{20} < 1

and

y^{2} < 5 x

.

Solving the boundary equations:

\frac{x ^{2}}{8} + \frac{y ^{2}}{20} = 1

and

y^{2} = 5 x

gives

x = 2, y = \pm 10

.

The integer points in X are:

X = {(1, 1), (1, 0), (1, - 1), (1, 2), (1, - 2), (2, 3), (2, 2), (2, 1), (2, 0), (2, - 1), (2, - 2), (2, - 3)}

Total ways to choose 3 points:

(3 12) = 220

.

For the area to be a positive integer, we need triangles with integer area. The area formula is

\frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣

.

Triangles with base 2 (between

x = 1

and

x = 2

with even vertical distance):

3 \times 7 + 5 \times 5 = 46

Triangles with base 4:

1 \times 7 + 3 \times 5 = 22

Triangles with base 6:

1 \times 5 = 5

Total favorable outcomes:

46 + 22 + 5 = 73

Probability

= \frac{73}{220}

Answer: (B)

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Single correct option

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JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Probability

Solution

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