In the first three articles in this series on QUANT probability questions, I discussed the AND and OR probability rules, "**at least**" probability questions, and probability questions that involve counting. This post covers relatively rare kind of probability question not covered in the first three posts: **geometric probability** questions. First, a few challenging practice QUANT questions on this topic.

1. In the diagram above, the sides of rectangle **not** inside the circle?

A.

B.

C.

D.

D.

2. Region

A.

B.

C.

D.

E.

3. In the diagram above,

A.

B.

C.

D.

E.

4. If points

A.

B.

C.

D.

E.

These three are very challenging problems. If they seem easy to you, you are probably a QUANT Quantitative expert! If these make your head spin, please read on!

As anti-intuitive as this may seem, there is a deep mathematical link between probability and area. For example, if you are dealing with regions under the Normal Distribution or any other distribution, the probability of landing in those regions is measured by finding the area under the curve. This connection between area and probability because quite apparent if you happen to study calculus-based statistics. If you haven’t ventured into such regions, I’ll just say, remember: there’s an important link between probability and area. In particular, if a diagram is given or described in a Quantitative question, and you are asked for a probability, then you almost always will be finding a ratio of two areas. Thus, any geometric probability problem just becomes a 2-in1 find-the-area problem.

Know your basic geometry. You should know how to find the area of rectangles and triangles: if you don’t remember, say, the formula for the area of a trapezoid, just break it into triangles & rectangles and find the area piece-by-piece. You should know Archimedes' remarkable formula for the area of a circle:

You should also be comfortable with setting up the proportion necessary to find the area of a circular sector (i.e. a "**slice of the pie**").

If you are given a diagram as part of the problem, that’s great. If you are not given a diagram, sketch one on the notepad. In any problem involving Geometry, it’s very important to have your right-brain visual skills engaged.

The most general formula for probability is

In the case of geometry probability, each of these terms, the numerator and the denominator, is an area.

Many times in geometric probability problems, as in #1 and #3 above, no absolute lengths are given — we know the relative ratios, but we don’t know in absolute terms how big anything is. The information in the problem does not allow us to compute the numerical value of any area. That’s fine. Just pick **any convenient number** for one of the lengths, and figure out everything from there. It **doesn’t matter what value you pick**, because in the probability ratio, the value you pick will cancel. All that matters are the ratios.

With these recommendations, go back and look at those problems again before reading the explanations below.

1. We aren’t given any absolute lengths. For convenience, I am going to assume that the radius of the circle is **denominator area**". Now, for the numerator area, the area of the rectangle that does not include the circle, subtract the circle from the rectangle:

**Answer = D**

2. Here, we are not given a diagram, so we will sketch one. In the real test, just a rough diagram will be enough to visualize things. Here’s an accurate diagram.

Notice the square

Now, let’s think about this line. Let’s solve for

First of all, putting it in slope-intercept form makes clear — the

That’s the numerator-area, so just divide

**Answer = C**

3. Again, we are not given any absolute lengths, so we will pick something convenient. Here, I am going to pick

Both

Now, we can find the area of the circle, using Archimedes’ formula:

Because triangle

First of all, that’s the denominator-area in our probability. Also, we can use this to figure out the area of the shaded region.

That’s our numerator-area. Now, we can put together the probability — we will just have to simply the complex fraction by multiplying numerator and denominator by

**Answer = A**

4. First of all, the circle has what is called rotational symmetry, and we will take advantage of this. Because of rotational symmetry, we can pick, at random, any location we want for point

Because we are interested in an inequality (chord **strategy of solving the equation first** (chord

Also, keep in mind — because point B could be randomly located anywhere on the circle, it could be on either side of point

In that diagram, chords

Notice there's a most fortuitous set of connections in that diagram. There are five segments —

This means the two triangles — triangle **equilateral triangles**. (In fact, any time you have

There are

If point

**Answer = D**