Question

Consider the 6 x 6 square in the figure. Let A1, A2, ........, A49 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 a column. Assume that each point A_i has an equal chance of being chosen.  Let i p 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 7E(X) is

Answer 1

Probability and Expected Value Calculation 

Consider a 6x6 grid with 49 points \( A_1, A_2, \dots, A_{49} \). Two points \( A_i \) and \( A_j \) are friends if they are adjacent along a row or a column.

Step 1: Counting the Number of Friends

- **Corner Points** (4 points): Each has 2 friends.
- **Edge Points (excluding corners)** (20 points): Each has 3 friends.
- **Internal Points** (25 points): Each has 4 friends.

Step 2: Probabilities \( p_i \)

- The probability \( p_0 \) that a point has 0 friends is \( p_0 = 0 \).
- The probability \( p_1 \) that a point has 1 friend is \( p_1 = 0 \).
- The probability \( p_2 \) that a point has 2 friends is: \[ p_2 = \frac{4}{49} \] - The probability \( p_3 \) that a point has 3 friends is: \[ p_3 = \frac{20}{49} \] - The probability \( p_4 \) that a point has 4 friends is: \[ p_4 = \frac{25}{49} \]

Step 3: Expected Value \( E(X) \)

The expected value \( E(X) \) is: \[ E(X) = 2 \cdot \frac{4}{49} + 3 \cdot \frac{20}{49} + 4 \cdot \frac{25}{49} = \frac{24}{7} \]

Step 4: Calculating \( 7E(X) \)

The value of \( 7E(X) \) is: \[ 7E(X) = 7 \cdot \frac{24}{7} = 24 \]

Final Answer

The value of \( 7E(X) \) is \( \boxed{24} \).