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 and A are friends if they are adjacent along a row or along a column. Assume that each point A has an equal chance of being chosen.

Two distinct points are chosen randomly out of the points A₁, A₂, . . . , A₄₉. Let p be the probability that they are friends. Then the value of 7p is:

The correct answer is 0.5.

Step 1: Understand the total number of points and total ways to choose two distinct points.
The given grid is a 6 × 6 square composed of small unit squares, which forms a grid of 7 vertical lines and 7 horizontal lines.
The points of intersection are denoted as A₁, A₂, ..., A₄₉. Thus, the total number of points is:

$N=7\times 7=49$

The total number of ways to choose 2 distinct points out of 49 points is given by the combination formula ⁴⁹C₂:

$\text{Total outcomes}=\left(\genfrac{}{}{0ex}{}{49}{2}\right)=\frac{49\times 48}{2}=49\times 24=1176$

Step 2: Count the total number of friendly pairs.
Two distinct points are considered "friends" if they are directly adjacent along a horizontal row or along a vertical column.
Each pair of adjacent points corresponds to a single line segment of length 1 in the grid.

1. Horizontal friendly pairs:
There are 7 horizontal rows. Each row contains 7 points, which form 6 adjacent horizontal segments (pairs).

$\text{Total horizontal pairs}=7\times 6=42$

2. Vertical friendly pairs:
There are 7 vertical columns. Each column contains 7 points, which form 6 adjacent vertical segments (pairs).

$\text{Total vertical pairs}=7\times 6=42$

3. Total favorable pairs:

$\text{Favorable outcomes}=42+42=84$

Step 3: Calculate the probability p and the required value 7p.
The probability p that two randomly chosen distinct points are friends is:

$p=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}=\frac{84}{1176}=\frac{1}{14}$

Now, we find the value of 7p:

$7p=7\times \frac{1}{14}=\frac{1}{2}=0.5$