math.wiki
Probability › Basic probability › Geometric probability
Open on the map
Math 7 · 7.SP.C.7B

Geometric probability

A raindrop can land anywhere on a patio, and there are far too many landing spots to count. How likely is it to land in the drain?

You need
Measuring anglesGrades 4 to 8
Where you are 14 in Basic probability
Probability4
Basic probability14
▴ 1 aboveComplements and probabilities that add to oneDesigning a simulation
Geometric probability
Listing a sample space: lists and tablesOdds▾ Show all 14 (8 more)

When you reach for it

Counting outcomes needs a list of equally likely results, like the six faces of a die. A raindrop falling on a patio has no such list: it can come down at any point, and there are more points than any count could reach. So can a hailstone striking a flat roof, a parachutist coming down in a field, or a ball rolled across a floor until it stops.

Reach for area when the outcome is a position inside a region and nothing favors one part of the region over another. Then the chance of landing in a part is that part’s share of the area. A spinner is the same idea on a disc, where each slice’s share is set by its angle.

The procedure

What “at random in a region” means

A point lands at random in a region when it is certain to land somewhere in the region, and any two parts with the same area are equally likely to receive it, whatever their shape and wherever they sit. That is a claim about the situation, not something the mathematics promises. Rain falling evenly on a small patio fits it; a parachutist steering for the middle of a field does not.

Why probability becomes area

Cover the whole region with a grid of small equal squares. In the figure the region is a 6 by 4 rectangle, cut into 24 unit squares, and the shaded target covers 6 of them.

6
4
Figure 1Equal squares are equally likely, so the shaded target’s chance is the share of squares it covers, \displaystyle \frac{6}{24} = \frac{1}{4}, which is its share of the area.

Equal squares are equally likely, and exactly one of them receives the point, so the counting rule applies:

\begin{aligned} P(\text{target}) &= \frac{\text{squares in the target}}{\text{squares in the whole}} \\ &= \frac{6}{24} = \frac{1}{4} \end{aligned}

If each square has area a, the target holds \displaystyle \frac{\text{target area}}{a} squares and the whole holds \displaystyle \frac{\text{whole area}}{a}. In the ratio the a cancels. So, for a point that lands at random, with every spot of the region equally likely,

P(\text{target}) = \frac{\text{area of the target}}{\text{area of the whole}}

Without that assumption the squares are no longer equally likely, the counting rule does not apply, and neither does this ratio.

A target with a curved edge, such as a circle, cuts through some squares. Make the grid finer and the cut squares form a thinner band along the edge, a smaller share of the count, so the ratio of counts settles on the ratio of the true areas.

Using the rule

The whole region is everything that can happen, and its area goes on the bottom. For a raindrop known to have landed on the patio, that is the patio.

The target lies inside the whole, and its area goes on top. Find it with the formulas for rectangles, triangles and circles, or by splitting and subtracting. Since the target is part of the whole, the probability lies between 0 and 1.

Both areas must be in the same unit. Square centimeters over square meters means nothing, so convert first: 1\ \text{m}^2 = 100 \times 100 = 10{,}000\ \text{cm}^2 (converting square units). Once the units match they cancel, and the probability has no unit.

The fraction is exact. The usual conversions give a decimal or percent. When a circle is involved, keep \pi in the exact answer and round only for the decimal.

Missing the target is the complement. The point lands in the target or in the rest, never both, so by the complement rule

\begin{aligned} P(\text{miss}) &= 1 - P(\text{target}) \\ &= \frac{\text{whole area} - \text{target area}}{\text{whole area}} \end{aligned}

Working backwards. The rule ties together the probability, the target area and the whole area, so any two give the third. Multiplying both sides by the whole area,

\text{target} = P \times \text{whole}

where both words stand for areas. That is how a target is designed to give a chosen probability.

Spinners

A spinner turns about a center and is equally likely to stop facing any direction. Its parts are sectors, slices cut by two radii with an angle of \theta degrees between them.

Cut the disc into 360 slices of 1^\circ. Turning the disc carries any slice onto any other, so all have the same area, each \displaystyle \frac{1}{360} of the disc. A sector of \theta^\circ holds \theta of them, so its area is \displaystyle \frac{\theta}{360}\pi r^2 (the area of a sector), and

P(\text{sector}) = \frac{\frac{\theta}{360}\pi r^2}{\pi r^2} = \frac{\theta}{360}

The \pi r^2 cancels: the size of the wheel does not matter, only the angle at the center, measured in degrees. Probability models reached the same share by thinking about directions; area gives it again. The sectors fill the wheel, so their angles add to 360^\circ and their probabilities to 1.

Worked examples

Example 1

A hailstone on a roof with a skylight

A hailstone lands at a random spot on a flat rectangular roof 8 m long and 5 m wide. The roof holds a rectangular skylight 3 m long and 2 m wide. Find the probability that the hailstone lands on the skylight, and that it lands on the roof but not the skylight, each as a fraction, a decimal and a percent.

Solution

The hailstone is known to land on the roof, so the roof is the whole: 8 \times 5 = 40\ \text{m}^2. The skylight is the target: 3 \times 2 = 6\ \text{m}^2. Both are in square meters.

P(\text{skylight}) = \frac{6}{40} = \frac{3}{20} = 0.15 = 15\%

Missing the skylight is the complement:

P(\text{not the skylight}) = 1 - \frac{3}{20} = \frac{17}{20} = 0.85 = 85\%

Check: the roof outside the skylight is 40 - 6 = 34\ \text{m}^2, and \displaystyle \frac{34}{40} = \frac{17}{20}. Where the skylight sits does not matter, only its area.

Example 2

A raindrop and a square drain

A patio is a rectangle 5 m by 4 m, with a square drain 40 cm on each side set into it. A raindrop lands at a random spot on the patio. Find the probability that it lands on the drain. If 2{,}500 drops land on the patio, about how many fall on the drain?

Solution

The units differ, so convert first. The drain’s side is 0.4 m, so its area is 0.4 \times 0.4 = 0.16\ \text{m}^2, and the patio’s area is 5 \times 4 = 20\ \text{m}^2.

P(\text{drain}) = \frac{0.16}{20} = 0.008 = 0.8\%

As a fraction, \displaystyle 0.008 = \frac{1}{125}. Check in centimeters: \displaystyle \frac{40 \times 40}{500 \times 400} = \frac{1{,}600}{200{,}000} = \frac{1}{125}.

A model with probability p predicts about pN hits in N trials (probability models), so expect about

2{,}500 \times \frac{1}{125} = 20

drops on the drain: where the count is likely to sit, not what it will be.

Example 3

A parachutist and a target circle

A parachutist comes down at a random spot in a rectangular field 50 m long and 40 m wide. A target circle of radius 10 m is painted wholly inside the field. Find the probability of landing inside the circle, exactly and to three decimal places, and of landing outside it.

Solution

The field is the whole: 50 \times 40 = 2{,}000\ \text{m}^2. The circle’s area is \pi \times 10^2 = 100\pi\ \text{m}^2.

P(\text{circle}) = \frac{100\pi}{2{,}000} = \frac{\pi}{20}

That is exact. As a decimal, \displaystyle \frac{\pi}{20} \approx 0.157, about 15.7\%. Landing outside is the complement:

P(\text{outside}) = 1 - \frac{\pi}{20} \approx 0.843

“Wholly inside” matters: the circle is 20 m across, less than the field’s width, so it can sit inside, and then all of its area belongs to the field.

Example 4

A button on a patchwork quilt

A quilt 2.4 m by 1.6 m is sewn from square patches 40 cm on a side. Each patch has a dark right triangle whose shorter sides run along the whole 40 cm bottom edge and 30 cm up its left edge. A button dropped on the spread-out quilt lands at a random spot. Find the probability that it lands on a dark triangle.

Solution

One patch has area 40 \times 40 = 1{,}600\ \text{cm}^2. The triangle’s shorter sides meet at a right angle, so they serve as base and height: \displaystyle \frac{1}{2} \times 40 \times 30 = 600\ \text{cm}^2.

The quilt is 240 cm by 160 cm, so it holds 6 patches across and 4 down, 24 patches in all. The 24 cancels from top and bottom:

P(\text{dark}) = \frac{24 \times 600}{24 \times 1{,}600} = \frac{3}{8} = 0.375 = 37.5\%

One patch alone gives the same answer, however many patches there are.

Example 5

A game wheel with four sectors

A game wheel of radius 20 cm has sectors A, B, C and D with angles 45^\circ, 75^\circ, 100^\circ and 140^\circ. It is spun hard, so it is equally likely to stop at any position. Find the probability of each sector, and of stopping on A or B.

45^\circ
 A
75^\circ
 B
100^\circ
 C
140^\circ
 D
Figure 2The game wheel in Example 5. The sector angles add to 360^\circ.
Solution

The sectors fill the wheel, since 45 + 75 + 100 + 140 = 360. By area, sector A is \displaystyle \frac{45}{360} \times \pi \times 20^2 = 50\pi\ \text{cm}^2 of a wheel of 400\pi\ \text{cm}^2, so \displaystyle P(\text{A}) = \frac{50\pi}{400\pi} = \frac{1}{8}. The radius and \pi cancel, as the spinner rule says, so the rest need only their angles:

SectorProbability
A\displaystyle \frac{45}{360} = \frac{1}{8}
B\displaystyle \frac{75}{360} = \frac{5}{24}
C\displaystyle \frac{100}{360} = \frac{5}{18}
D\displaystyle \frac{140}{360} = \frac{7}{18}

As decimals these are 0.125, about 0.208, about 0.278 and about 0.389.

Check over the denominator 72: \displaystyle \frac{9}{72} + \frac{15}{72} + \frac{20}{72} + \frac{28}{72} = 1.

A and B sit side by side and make one sector of 45 + 75 = 120 degrees, so

P(\text{A or B}) = \frac{120}{360} = \frac{1}{3}

which agrees with \displaystyle \frac{1}{8} + \frac{5}{24} = \frac{8}{24}, since the wheel cannot stop on both.

Example 6

Designing for a chosen probability

(a) Achille wants the jackpot sector of his game wheel to come up with probability 0.15. What angle should it have?

(b) Ingrid’s flat roof is 10 m by 9 m. She is fitting a rectangular skylight 2 m wide and wants a hailstone landing at random on the roof to hit it with probability \displaystyle \frac{1}{15}. How long should the skylight be?

Solution

Both run the rule backwards: the probability is known, and part of the target is not.

(a) A sector’s probability is \displaystyle \frac{\theta}{360}, so

\begin{aligned} \frac{\theta}{360} &= 0.15 \\ \theta &= 0.15 \times 360 = 54 \end{aligned}

The jackpot needs 54^\circ. Check: \displaystyle \frac{54}{360} = 0.15.

(b) The roof’s area is 10 \times 9 = 90\ \text{m}^2, so the skylight must have area \displaystyle \frac{1}{15} \times 90 = 6\ \text{m}^2. Its length \ell satisfies 2\ell = 6, so \ell = 3 m, and a 2 m by 3 m skylight fits on the roof. Check: \displaystyle \frac{6}{90} = \frac{1}{15}.

Boundaries

Every spot must be equally likely. A throw aimed at the middle of a target, wind pushing rain or hail toward one side, a parachutist steering, or a wheel that sticks all break the model, and area then gives the wrong answer. Such a probability has to come from recorded trials instead.

The point must be certain to land in the whole region. If a drop might miss the patio, the patio is not the whole region.

Only the part of a target inside the whole counts. A circle hanging over the edge of a field contributes only the area inside the field, and where two targets overlap, the shared part is counted once.

Both areas in one unit, or the ratio means nothing.

A line has no area. The edge of a drain or the boundary between two sectors has area 0, so landing exactly on it has probability 0, and whether an edge counts as inside never changes an answer. Probability 0 here does not mean impossible: the point does land somewhere, and every single point has probability 0.

The angle rule needs a slice from the center. \displaystyle \frac{\theta}{360} holds for a sector cut by two radii, on a wheel equally likely to stop facing any direction. Any other region of a wheel, such as a ring near the rim, needs its area instead.

Where this goes next

Letting area take over from counting is the step from lists of outcomes to measured quantities. A time or a length can take any value in a range, as the raindrop can land at any point, and density curves carry the idea on: probability becomes area under a curve, with the whole area equal to 1.

The rule also runs the other way. Drop many random points into a square holding a circle that just fits, and the share landing inside estimates the circle’s share of the area, \displaystyle \frac{\pi}{4}. That is a simulation, and the law of large numbers explains why the estimate improves as the points pile up.

Exercises

Rectangles and triangles

  1. A point is chosen at random inside a rectangle 10 cm by 8 cm. A square of side 4 cm lies inside it. Find the probability that the point lands in the square, as a fraction, a decimal and a percent.
  2. A hailstone lands at a random spot on a flat roof 9 m by 6 m, which carries a solar panel 4.5 m by 2 m. Find the probability that it lands (a) on the panel, (b) not on the panel.
  3. A patio is 6 m by 5 m, with a square drain 30 cm on a side. Find the probability that a raindrop landing at random on the patio lands on the drain, as a decimal and a percent.
  4. A point is chosen at random inside a rectangle 12 cm long and 8 cm tall. A triangle inside it has a base of 6 cm along the bottom side and its third corner on the top side. Find the probability that the point lands in the triangle.
  5. A tablecloth is printed all over with squares 25 cm on a side. Each square has a shaded triangle with a base of 25 cm along one edge and a height of 16 cm. A crumb falls at a random spot on the cloth. Find the probability that it lands on a shaded triangle.

Circles and combined regions

  1. A parachutist lands at a random spot in a field 100 m by 60 m holding a target circle of radius 6 m. Find the probability of landing in the circle, exactly and as a percent to one decimal place.
  2. A circle of radius 3 cm is drawn inside a square of side 10 cm. Find the probability that a point chosen at random in the square lies inside the circle, exactly and to three decimal places.
  3. A floor 6 m by 5 m has a circular rug 2 m across on it. A ball stops at a random spot on the floor. Find the probability, to three decimal places, that it stops (a) on the rug, (b) off the rug.
  4. A patio 6 m by 4 m has a square drain 50 cm on a side and a round planter of radius 1 m, not overlapping. Find the probability, to three decimal places, that a raindrop landing at random on the patio misses both.

Spinners

  1. For the game wheel of Example 5, find the probability that it stops (a) on B or C, (b) on anything but D.
  2. A spinner has sectors of 30^\circ, 110^\circ and 220^\circ. Check that they fill the spinner, and find each probability as a fraction and as a decimal to three places.
  3. A wheel has four sectors, three of them 80^\circ, 95^\circ and 120^\circ. Find the angle of the fourth and the probability of stopping on it.

Working backwards

  1. What angle must a sector have for a spinner to stop on it with probability 0.35?
  2. A flat roof is 15 m by 8 m, and a rectangular solar panel on it is to be 5 m long. How wide should the panel be for a hailstone landing at random on the roof to hit it with probability \displaystyle \frac{1}{12}?
  3. A rectangular field is 40 m wide. A target circle of area 50\ \text{m}^2 lies inside it, and a parachutist landing at random in the field lands in the circle with probability 0.025. How long is the field?

Reasoning

  1. Dervla says a raindrop is less likely to land on a 1\ \text{m}^2 patch at a corner of a patio than on a 1\ \text{m}^2 patch in the middle. Under this lesson’s model, is she right? Explain.
  2. Two wheels, of radius 10 cm and 30 cm, each have a gold sector of 45^\circ. Explain why gold is equally likely on both, using the areas of the sectors.
  3. Erasmus says the model in Example 3 is poor for a skilled parachutist. Explain what assumption fails, and which way it pushes the real chance of landing in the target.
  4. Explain why the probability of a raindrop landing exactly on the edge of the drain is 0 under the model.
  5. Fenwick’s target circle hangs partly over the edge of a field, and he works out the probability as the whole circle’s area divided by the field’s area. Explain what is wrong.
Answers
  1. \displaystyle \frac{1}{5} = 0.2 = 20\%. The square is 16\ \text{cm}^2 and the rectangle 80\ \text{cm}^2.
  2. (a) \displaystyle \frac{1}{6} \approx 0.167. The panel is 9\ \text{m}^2 and the roof 54\ \text{m}^2. (b) \displaystyle \frac{5}{6} \approx 0.833, by the complement rule.
  3. 0.003 = 0.3\%. The drain is 0.3 \times 0.3 = 0.09\ \text{m}^2 and the patio 30\ \text{m}^2.
  4. \displaystyle \frac{1}{4}. Wherever the third corner sits on the top side, the height is 8 cm, so the area is \displaystyle \frac{1}{2} \times 6 \times 8 = 24\ \text{cm}^2, out of 96\ \text{cm}^2.
  5. \displaystyle \frac{8}{25} = 0.32 = 32\%. One triangle is \displaystyle \frac{1}{2} \times 25 \times 16 = 200\ \text{cm}^2 of a 625\ \text{cm}^2 square, and every square is alike.
  6. \displaystyle \frac{3\pi}{500} \approx 0.0188, about 1.9\%. The circle is 36\pi\ \text{m}^2 and the field 6{,}000\ \text{m}^2.
  7. \displaystyle \frac{9\pi}{100} \approx 0.283. The circle is 9\pi\ \text{cm}^2 and the square 100\ \text{cm}^2.
  8. (a) \displaystyle \frac{\pi}{30} \approx 0.105. The rug has radius 1 m, so area \pi\ \text{m}^2, and the floor is 30\ \text{m}^2. (b) \displaystyle 1 - \frac{\pi}{30} \approx 0.895.
  9. \displaystyle \frac{23.75 - \pi}{24} \approx 0.859. The patio is 24\ \text{m}^2, the drain 0.25\ \text{m}^2 and the planter \pi\ \text{m}^2, leaving 23.75 - \pi\ \text{m}^2 bare.
  10. (a) \displaystyle \frac{35}{72} \approx 0.486, from \displaystyle \frac{75 + 100}{360}. (b) \displaystyle \frac{11}{18} \approx 0.611, from \displaystyle \frac{360 - 140}{360}.
  11. 30 + 110 + 220 = 360. The probabilities are \displaystyle \frac{1}{12} \approx 0.083, \displaystyle \frac{11}{36} \approx 0.306 and \displaystyle \frac{11}{18} \approx 0.611.
  12. 65^\circ, from 360 - 295, with probability \displaystyle \frac{65}{360} = \frac{13}{72} \approx 0.181.
  13. 126^\circ. \theta = 0.35 \times 360 = 126.
  14. 2 m. The panel must be \displaystyle \frac{1}{12} \times 120 = 10\ \text{m}^2, and 10 \div 5 = 2.
  15. 50 m. The field’s area is 50 \div 0.025 = 2{,}000\ \text{m}^2, and 2{,}000 \div 40 = 50.
  16. No. Under the model, parts with equal areas are equally likely wherever they sit.
  17. The small sector is \displaystyle \frac{45}{360} \times 100\pi = 12.5\pi\ \text{cm}^2 of 100\pi, and the large one \displaystyle \frac{45}{360} \times 900\pi = 112.5\pi\ \text{cm}^2 of 900\pi. Both shares are \displaystyle \frac{1}{8}.
  18. The model assumes every spot is equally likely. A skilled parachutist steers for the target, so the real chance of landing in it is well above \displaystyle \frac{\pi}{20}.
  19. The edge is a line, with area 0, and 0 divided by the patio’s area is 0.
  20. Only the part of the circle inside the field can be landed on, so only that part’s area belongs on top. The overhanging part adds area no landing can reach, so his answer is too big.
kg1128 · updated 2026-10-08 · CC-BY-SA-4.0