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Expectation of Friends Count in Square Grid

Comprehension Passage

Consider the 6×66 \times 6 square in the figure. Let A1,A2,,A49A_1, A_2, \dots, A_{49} be the points of intersections (dots in the picture) in some order. We say that AiA_i and AjA_j are friends if they are adjacent along a row or along a column. Assume that each point AiA_i has an equal chance of being chosen.

Let pip_i be the probability that a randomly chosen point has ii many friends, i=0,1,2,3,4i = 0, 1, 2, 3, 4. Let XX be a random variable such that for i=0,1,2,3,4i = 0, 1, 2, 3, 4, the probability P(X=i)=piP(X = i) = p_i. Then the value of 7E(X)7E(X) is

Question Diagram 1
Official Numerical Answer24

Step-by-Step Solution

To find the value of 7E(X)7E(X), we analyze the grid of intersection points.

1. Understanding the Grid Structure

The given 6×66 \times 6 square grid contains 77 horizontal lines and 77 vertical lines. The total number of intersection points (vertices) is: Total points=7×7=49\text{Total points} = 7 \times 7 = 49

Two points AiA_i and AjA_j are "friends" if they are adjacent vertically or horizontally. Thus, the number of friends of a point corresponds to its degree in the grid graph.


2. Counting Points by Degree (Number of Friends)

The 4949 points can be categorized based on their position in the grid:

  1. Corner Points (i=2i = 2 friends):

    • There are 44 corners.
    • Each corner point is connected to 22 adjacent points (one vertically, one horizontally).
    • Number of points with 22 friends = 44.
  2. Boundary Non-Corner Points (i=3i = 3 friends):

    • Each of the 44 outer sides has 72=57 - 2 = 5 non-corner points.
    • Total boundary non-corner points = 4×5=204 \times 5 = 20.
    • Each such point is connected to 33 adjacent points.
    • Number of points with 33 friends = 2020.
  3. Interior Points (i=4i = 4 friends):

    • The interior forms a (72)×(72)=5×5(7 - 2) \times (7 - 2) = 5 \times 5 sub-grid.
    • Total interior points = 2525.
    • Each interior point is connected to 44 adjacent points (left, right, up, down).
    • Number of points with 44 friends = 2525.
  4. Points with i=0i = 0 or i=1i = 1 friends:

    • There are no isolated or end-degree points in the grid.
    • Number of points with 00 or 11 friends = 00.

3. Probability Distribution and Expectation Calculation

The probability pi=P(X=i)p_i = P(X = i) for each ii is given by: p0=0p_0 = 0 p1=0p_1 = 0 p2=449p_2 = \frac{4}{49} p3=2049p_3 = \frac{20}{49} p4=2549p_4 = \frac{25}{49}

The expected value E(X)E(X) of the random variable XX is: E(X)=i=04ipiE(X) = \sum_{i=0}^4 i \cdot p_i E(X)=0(0)+1(0)+2(449)+3(2049)+4(2549)E(X) = 0(0) + 1(0) + 2\left(\frac{4}{49}\right) + 3\left(\frac{20}{49}\right) + 4\left(\frac{25}{49}\right) E(X)=8+60+10049=16849=247E(X) = \frac{8 + 60 + 100}{49} = \frac{168}{49} = \frac{24}{7}


4. Alternative Method (Sum of Degrees)

Alternatively, by graph theory (Handshaking Lemma):

  • Total horizontal edges = 7×6=427 \times 6 = 42
  • Total vertical edges = 7×6=427 \times 6 = 42
  • Total edges E=42+42=84E = 42 + 42 = 84

The sum of degrees of all 49 points is 2E=2×84=1682E = 2 \times 84 = 168.

Thus, the average degree (expected value E(X)E(X)) is: E(X)=Sum of degreesTotal points=16849=247E(X) = \frac{\text{Sum of degrees}}{\text{Total points}} = \frac{168}{49} = \frac{24}{7}


Conclusion

We need to find the value of 7E(X)7E(X): 7E(X)=7×247=247E(X) = 7 \times \frac{24}{7} = 24

Expectation of Friends Count in Square Grid | Mathematics PYQ Solution - JEE Challenger