Marcus Bass

2022-09-26

Geometric probability

Inside a square of side 2 units , five points are marked at random. What is the probability that there are at least two points such that the distance between them is at most $\sqrt{2}$ units?

Inside a square of side 2 units , five points are marked at random. What is the probability that there are at least two points such that the distance between them is at most $\sqrt{2}$ units?

Matthias Calhoun

Beginner2022-09-27Added 11 answers

Step 1

$p=0$ because the furthest separation between 5 points in a square is with one at each corner and one in the middle, but by Pythagoras theorem the distance between the corner and the middle is $\sqrt{2}$.

Step 2

Not a formal proof I grant you but the logic is infallible.

$p=0$ because the furthest separation between 5 points in a square is with one at each corner and one in the middle, but by Pythagoras theorem the distance between the corner and the middle is $\sqrt{2}$.

Step 2

Not a formal proof I grant you but the logic is infallible.

Kelton Molina

Beginner2022-09-28Added 1 answers

Step 1

Divide the square into 4 squares of side length 1.In at least one square there are two or more points.

Step 2

All points in the same square have distance less than ${2}^{0.5}$. So the answer is 1

Divide the square into 4 squares of side length 1.In at least one square there are two or more points.

Step 2

All points in the same square have distance less than ${2}^{0.5}$. So the answer is 1

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