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Uniform models on a segment or a region, where probability is measure over measure, and probability zero is not impossible.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
You will build a uniform model on a segment or a region, describe the favourable set, measure it and the sample space in the same units, and divide to get a probability. You will also say what probability zero means when the outcome is a point, and use the complement when the favourable region is the awkward one to measure.
Everything in this unit so far divided one count by another. A point chosen on a rod has infinitely many possible positions and no two of them can be counted, so the equally likely model has to be rebuilt on something other than counting. Length is the something.
A model is uniform on a set when no part of it is favoured over another part of the same size. Measure is the general word for length, area or volume, whichever the setting calls for. The support of a model is the set over which its measure is positive; outside it, no probability accumulates at all.
When an outcome is a point of an interval, a region or a solid, and no part of it is favoured over any other of the same size, the model is uniform and $$P(A) = \frac{\text{measure of } A}{\text{measure of } \Omega},$$ where measure is length, area or volume as the setting demands.
This satisfies the axioms for exactly the reason area does: it is never negative, the whole space gets $1$, and disjoint pieces add. Nothing new is being assumed about probability; a different measure is being used to supply it.
Two consequences are worth stating out loud, because both contradict the finite intuition. A single outcome has probability zero — a point has no length — and it is nonetheless a possible outcome, since the break lands somewhere. And the endpoints of an interval do not matter: $[a, b]$ and $(a, b)$ have the same length, so they have the same probability, which is why a continuous model is indifferent to strict and non-strict inequalities in a way a discrete one is not.
Another way: picture
A bus arrives at a time chosen with no preference in a ten-minute window, drawn as a ten-centimetre strip of paper. The probability it comes in the first three minutes is the first three centimetres of the strip: $0.3$, measured with a ruler rather than counted.
Another way: steps
Two quantities chosen independently and uniformly make a square of outcomes, and a condition relating them cuts a region out of it.
| Condition on $(x, y)$ in the unit square | Favourable region | Area |
|---|---|---|
| $x < y$ | one triangle | $1/2$ |
| $x + y < 1$ | one triangle | $1/2$ |
| $\|x - y\| < t$ | a band along the diagonal | $1 - (1-t)^{2}$ |
| $xy < 1/4$ | the region under a hyperbola | needs an integral |
The third line is the meeting problem, and it is the model case for measuring the complement: the band is awkward and the two corners it leaves are triangles.
Reading probability zero as impossible. In a continuous model every single outcome has probability zero and one of them happens. The right reading is negligible in this measure.
Counting marks instead of measuring. A rod does not have twenty outcomes because it has twenty centimetre marks. The marks are a way of describing positions, not a list of them.
Fussing over endpoints. Whether the interval is open or closed changes no probability here. It matters for describing the set, which is why the set-building practice in this lesson asks for it, and not for the number.
Measuring in the wrong dimension. If the outcome is a pair of numbers, the sample space is a region and the favourable set has an area. Treating one of the two coordinates as the whole problem is the commonest way the meeting problem is got wrong.
A point is chosen with no preference on the segment from $0$ to $12$. What is the probability it exceeds $9$?
Sample space: the segment, length $12$.
The favourable set is the segment from $9$ to $12$, of length $3$.
Shade and measure.
So the probability is $3/12 = 0.25$.
Measure over measure.
$x$ and $y$ are each chosen with no preference in $[0, 1]$, independently. What is the probability that $x + y < 1$?
Sample space: the unit square, area $1$.
The favourable set is the triangle below the line $x + y = 1$, of area $1/2$.
Draw the line and shade beneath it.
So the probability is $1/2$ — and notice that no integral was needed, only the area of a triangle.
Geometry, not calculus, whenever the region is a polygon.
The favourable set is two pieces: from $0$ to $2$ and from $8$ to $10$.
Shade both; they are disjoint.
Their total length is $2 + 2 = 4$.
Additivity over disjoint pieces.
So the probability is $4/10 = 0.4$.
A rod $20$ centimetres long is snapped at a point chosen with no preference for any part of it. Give the probability that the break falls in each stretch.
| Probability | |
|---|---|
| Within $6$ cm of the left end | |
| Between $6$ cm and $14$ cm from the left end | |
| More than $14$ cm from the left end |
The same $20$ centimetre rod is snapped at a point chosen with no preference. Give the length of the stretch in which both resulting pieces are at least $3$ centimetres long, and the probability that the break falls there.
Favourable length: p cm. Probability: q.
A rod $20$ centimetres long is snapped at a point $x$ centimetres from the left end. Give the set of values of $x$ for which both pieces are at least $4$ centimetres long, as an interval.
This task has no paper form; do it on a device.
The rod is snapped at a point chosen with no preference. What is the probability that the break is at exactly $7$ centimetres from the left end, and what does that number mean?
Two people each arrive at a meeting point at a time chosen with no preference in the same hour, independently of each other, and each waits $6$ minutes before leaving. What is the probability that they meet?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
The $20$ centimetre rod is snapped at a point chosen with no preference. The marks below are one twentieth apart. Put the marker at the probability that the break falls between $6$ and $12$ centimetres from the left end.
0 |——————————| 1
Mark the position with a cross, then write the value:
You can set up a uniform model on a segment or in a square, describe and measure a favourable set, and switch to the complement when that is easier. Say in your own words why a single break point has probability zero without being impossible.
9. Your turn: a point is chosen with no preference on the segment from $0$ to $10$; find the probability it is within $2$ of an end, step 3