JEE Advanced 2023 — Paper 2 — MATH-IV solution

JEE Advanced2023Paper 2MATH-IV

PARAGRAPH “II”

Consider the $6 \times 6$ square in the figure. Let $A_{1}, A_{2}, \dots, A_{49}$ be the points of intersections (dots in the picture) in some order. We say that $A_{i}$ and $A_{j}$ are friends if they are adjacent along a row or along a column. Assume that each point $A_{i}$ has an equal chance of being chosen.

Q1. Let $p_{i}$ be the probability that a randomly chosen point has $i$ many friends, $i = 0, 1, 2, 3, 4$ . Let $X$ be a random variable such that for $i = 0, 1, 2, 3, 4$ , the probability $P (X = i) = p_{i}$ . Then the value of $7 E (X)$ is

Your answer

Solution

The lattice has $7 \times 7 = 49$ points. Classify by number of grid neighbours:

Corner points (exactly $2$ neighbours): $4$ corners, so $p_{2} = \frac{4}{49}$ .
Non-corner edge points (exactly $3$ neighbours): each side has $7 - 2 = 5$ such points, total $4 \times 5 = 20$ , so $p_{3} = \frac{20}{49}$ .
Interior points (exactly $4$ neighbours): $5 \times 5 = 25$ , so $p_{4} = \frac{25}{49}$ .
$p_{0} = p_{1} = 0$ since every lattice point has at least $2$ neighbours.

$E (X) = 2 \cdot \frac{4}{49} + 3 \cdot \frac{20}{49} + 4 \cdot \frac{25}{49} = \frac{8 + 60 + 100}{49} = \frac{168}{49} .$

Therefore $7 E (X) = \frac{168}{49} \cdot 7 = \frac{168}{7} = 24.$

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