Triangle from a square

Probability Level 1

Three points are chosen uniformly at random along the edges of a square. What is the probability a triangle can be formed from these points?

Clarification : The triangle formed should be non-degenerate , meaning the three points are not collinear.

\frac{1}{2}

\frac{2}{3}

\frac{3}{4}

\frac{7}{8}

\frac{15}{16}

1 solution

Andy Hayes
Oct 4, 2017

A triangle can be formed as long as all three points are not on the same edge.

Consider the probability that all three points are on the same edge. After the first point is selected, there is a $\frac{1}{4}$ probability that the second point is along the same edge, and then there is a $\frac{1}{4}$ probability again that the third point is also along the same egde. So the probability that all three points are along the same edge is $\left(\frac{1}{4}\right)\left(\frac{1}{4}\right)=\frac{1}{16}.$

The probability that a triangle can be formed is the complement of this probability, $1-\frac{1}{16}=\boxed{\frac{15}{16}}.$

But you can have 2 points on the same side and have a triangle..

Émilien Brais-Brunet - 3 years, 7 months ago

This solution includes triangles with 2 points on the same side of the square. $\frac{1}{16}$ is the probability that all three points are on the same side.

Andrew Hayes Staff - 3 years, 7 months ago

Friend.
PLs note you cant say all have same probability You should have mentioned this thing Also you have misses other possible ways

Janesh G - 3 years, 7 months ago

Whenever you see the phrase, "uniformly at random," that means that all choices are equally likely.

Andy Hayes - 3 years, 7 months ago

Hey just a doubt , why did you multiply 1/4 with 1/4 ?

Aman thegreat - 3 years, 8 months ago

That's probability rule of product . If there's a $p$ probability of one thing happening, and $q$ probability of another, independent thing happening, then there's a $p\times q$ probability of both of those things happening.

Andy Hayes - 3 years, 8 months ago

This is more about the question phrasing than the math but isn't there an edge case where two points coincide - the extreme end of one side and the coincident end of the perpendicular side (i.e.: which 'side' does a corner belong too?). Also I don't understand what 'uniformly at random' means.

Reg Migrant - 3 years, 7 months ago

"Uniformly at random" means that every single point is equally likely to be chosen. Since there are infinite points, the probability of any particular point being chosen is $0.$ So for your edge case, even though choosing coincident points or vertices is technically possible, the probability of that happening is $0.$

Andrew Hayes Staff - 3 years, 7 months ago

My answer: indeterminate. Number of favorable outcomes: infinite. Number of possible outcomes: infinite. Infinite/infinite= indeterminate.

Martin Gala - 3 years, 7 months ago

When dealing with infinities, an indeterminate form doesn't necessarily mean that there is no answer. It means the answer is indeterminate -- exactly that. In this case, the answer can be determined by noting that each side of the square is equally likely to be chosen.

Andrew Hayes Staff - 3 years, 7 months ago

Triangle from a square

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