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Video lessons › Mathematics · Unit 13 · Statistics & Probability · Chapter 2 · Classical Probability

Geometric Probability

8-minute video lesson in English, narrated with a computer voice.

Full notes, solved examples and practice for this lesson
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Geometric probability for JEE: when a point is chosen uniformly in a region, the probability is the favourable length, area or volume over the whole. A point in a square, a point in a circle (the radius trap), the band |x - y| ≤ a in a square, and two solved problems: two numbers that differ by at most 1/3, and the classic problem of two friends who wait 15 minutes for each other.

Chapters

  1. 0:00Introduction
  2. 0:43Probability by measuring, not counting
  3. 1:45Two quick checks: a square and a circle
  4. 3:10Two random numbers: the band picture
  5. 4:54Example 1: two numbers in 0 to 1 differ by at most 1/3
  6. 5:49Example 2: two friends wait 15 minutes, chance they meet
  7. 7:22Quick revision: area ratio and the band formula

Transcript

0:00Introduction

Hello everyone! Let us begin with a small puzzle. Two friends agree to meet somewhere between twelve and one o'clock. Each of them waits fifteen minutes for the other, and then leaves. So, what is the chance that they actually meet? Fifteen minutes out of sixty is one quarter, so the answer is one quarter, right? Hmm, not so fast. Each friend can arrive at any instant in that hour, so there are infinitely many possible times. And how do you count infinitely many outcomes? Today we stop counting, and start measuring. Questions like this do appear in JEE, so let's begin.

0:43Probability by measuring, not counting

So, here is the new setting. The outcomes are now points of a region, like a piece of a line, a square, or a circle. There are infinitely many points, so we cannot list and count them. And chosen uniformly means that no part of the region is preferred: two pieces of the same size are equally likely to contain the point. Okay? So let me ask you this. If we cannot count the points, what can we compare instead? Yes, exactly, we compare sizes! The probability of the event, written capital P with the event, capital A, in round brackets, is the favourable length, area or volume, over the whole length, area or volume. The favourable part, over the whole, that's all. Clear? Now, which size we use depends on the region. For points on a line, we use length. For points in a plane, we use area. And for points in space, we use volume. Today, all our questions live in a plane, so area is our main tool. Got it?

1:45Two quick checks: a square and a circle

Let us try two quick ones from the course. First, we choose a point at random in the unit square, the square with side one. Its coordinates, x and y, sit in the round brackets. We want P of x plus y less than one, and the less-than sign means strictly less than. So tell me, what fraction of the square has x plus y less than one? On the diagonal from the bottom right corner to the top left, x plus y is exactly one. Below it sits a triangle, and that triangle has exactly half the area of the square. The double arrow means that gives, so P equals one half. Right? Now let us take a circle. We choose a point at random inside a circle of radius two. What is the probability that it lies within distance one of the centre? One is half of two, so the answer is one half, isn't it? No, no, careful, that is the trap! The point is in a plane region, so we compare areas, not radii. The favourable part is a smaller circle of radius one, with area pi, the Greek letter, times one squared. The whole circle has area pi times two squared, and the small dot means times. So P is pi over four pi, which is one quarter, not one half. Area grows as the square of the radius. With me?

3:10Two random numbers: the band picture

Now the most useful picture of this topic. Two numbers, x and y, are chosen uniformly from zero up to capital L, okay? The square brackets mean the interval from zero to L, with both ends included. Here is the trick. Treat the pair as one point, x comma y. Then x is measured along the bottom, y up the side, and the point lands somewhere in this square of side L, so the whole area is L squared. The small two up there means squared. Clear? Often we want the two numbers to be close, within small a of each other. The vertical bars are the modulus, so mod x minus y is the gap between the two numbers, and the less-than sign with a line under it means less than or equal to. On the dashed diagonal, x equals y and the gap is zero, so the points near it form this shaded band. Right? So, how do we find the area of this slanted band? The smart move is to subtract the two white corners instead. In the top corner, y minus x is more than small a, and in the bottom corner, x minus y is more than small a. Each corner is a right triangle whose two legs are both L minus small a, so each has area one half times the square of L minus small a. Together, the two corners make L minus small a, the whole squared. Got it? So the band is L squared minus the corners, and we divide by the whole area, L squared. P of mod x minus y at most small a equals one minus the square of one minus small a over L, and the big round brackets hold the part being squared. Remember this line; it solves a whole family of questions. Okay?

4:54Example 1: two numbers in 0 to 1 differ by at most 1/3

Let's start with the easier example. Two numbers are chosen at random between zero and one. Find the probability that they differ by at most one third. Pause the video, and try it first. Done? Okay, let's check. Differ by at most one third means mod x minus y is at most one third, so this is exactly our band. Here L is one and small a is one third, so one minus small a over L is two thirds. That is the side of each corner triangle. With me? Now, two thirds squared is four ninths. So the answer is four ninths, right? Oh no, careful, four ninths is the area of the two white corners, the part we do not want! We want the band, so we subtract it from one. One minus four ninths gives five ninths. So the probability is five ninths, a little more than half. Clear?

5:49Example 2: two friends wait 15 minutes, chance they meet

Now, remember our puzzle from the start? Let's finish it. Two friends agree to meet between twelve and one o'clock, and each waits fifteen minutes for the other. Find the probability that they meet. Here is a hint: measure the time in hours. Pause the video, and try it first. Done? Okay, let's check. Let x and y be the arrival times of the two friends, in hours after twelve. Each one is a random number between zero and one, so the pair is a point in the unit square. With me? Now the key question. When exactly do the two friends meet? Yes, exactly, when they arrive within fifteen minutes of each other! Fifteen minutes, written min on screen, is one quarter of an hour. Whoever comes first waits, so they meet when mod x minus y is at most one quarter. It is our band again, with L equal to one and small a equal to one quarter. Right? So each corner has side one minus one quarter, which is three quarters. P equals one minus three quarters squared, that is one minus nine sixteenths, which gives seven sixteenths. See, the answer is not one quarter at all! Our quick guess used a single window of fifteen minutes. But either friend can come first, so the second friend may arrive up to fifteen minutes before or after the first, a window of thirty minutes, and the band covers both sides of the diagonal. Only near twelve and one o'clock is that window cut short, which is why the answer is a little less than one half, and much more than one quarter.

7:22Quick revision: area ratio and the band formula

Got it? So, quick revision. For a point chosen uniformly, P is the favourable length, area or volume, over the whole. Areas grow as the square of the size, so radius one inside radius two gives one quarter, not one half. And for two numbers between zero and L that differ by at most small a, P equals one minus the square of one minus small a over L, our band formula. Now go and try the practice problem on this page. You can do it. All the best!

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