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Probability from a uniform area model

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Geometric probability

When a point or object centre is uniformly distributed over a region, each unit of area is equally likely. Find the allowable region first, then use the favourable area divided by the total allowable area

\[\Pr(E)=\frac{A_E}{A_S}\]
  1. 1Find the allowable region For an object on a board, remove the object dimensions from the board dimensions \(\displaystyle w_c=13-4=9,\quad h_c=10-3=7\)
  2. 2Use area as the measure A uniform location over a two-dimensional region is measured by area, not by boundary length \(\displaystyle A=\text{width}\times\text{height}\)
  3. 3Calculate both areas The favourable area belongs in the numerator and the complete allowable area belongs in the denominator \(\displaystyle A_E=3\times2=6,\quad A_S=9\times7=63\)
  4. 4Keep the probability exact Reduce the area ratio and check that it is from zero to one \(\displaystyle \Pr(E)=\frac{6}{63}=\frac{2}{21}\)
Year 11, VCE Units 1 & 2, Chapter 8, 1 difficulty level, No calculatorBuild a worksheet

Geometric probability

When a point or object centre is uniformly distributed over a region, each unit of area is equally likely. Find the allowable region first, then use the favourable area divided by the total allowable area

\[\Pr(E)=\frac{A_E}{A_S}\]
  1. Find the allowable region For an object on a board, remove the object dimensions from the board dimensions \(\displaystyle w_c=13-4=9,\quad h_c=10-3=7\)
  2. Use area as the measure A uniform location over a two-dimensional region is measured by area, not by boundary length \(\displaystyle A=\text{width}\times\text{height}\)
  3. Calculate both areas The favourable area belongs in the numerator and the complete allowable area belongs in the denominator \(\displaystyle A_E=3\times2=6,\quad A_S=9\times7=63\)
  4. Keep the probability exact Reduce the area ratio and check that it is from zero to one \(\displaystyle \Pr(E)=\frac{6}{63}=\frac{2}{21}\)

Worked example

A rectangular object of width 4 units and height 3 units is placed on a board 16 units wide and 12 units high. Its centre is uniformly distributed over all positions that keep the object on the board. Event E occurs when the centre is in a target rectangle 6 units wide and 4 units high within the allowable centre region. Find Pr(E)

\[\Pr(E)\]
  1. \(\displaystyle w_S=16-4=12,\quad h_S=12-3=9\)Subtract the object dimensions to find the width and height available to its centre
  2. \(\displaystyle A_E=6\times4=24\)The target centre rectangle has area 24 square units
  3. \(\displaystyle A_S=12\times9=108\)The complete allowable centre region has area 108 square units
  4. \(\displaystyle \Pr(E)=\frac{A_E}{A_S}=\frac{24}{108}=\frac{2}{9}\)Divide the favourable area by the complete allowable area and reduce the fraction

Common mistakes

Using boundary length
✕\(\displaystyle \Pr(E)=\frac{2\left(4+3\right)}{2\left(10+8\right)}=\frac{7}{18}\)✓\(\displaystyle \Pr(E)=\frac{4\times3}{10\times8}=\frac{3}{20}\)
A uniform area model compares two-dimensional areas, not boundary lengths
Ignoring the clearance region
✕\(\displaystyle \Pr(E)=\frac{24}{16\times12}=\frac{1}{8}\)✓\(\displaystyle \Pr(E)=\frac{24}{\left(16-4\right)\left(12-3\right)}=\frac{2}{9}\)
The object centre cannot use positions where the object would extend beyond the board
Reversing the area ratio
✕\(\displaystyle \Pr(E)=\frac{80}{12}=\frac{20}{3}\)✓\(\displaystyle \Pr(E)=\frac{12}{80}=\frac{3}{20}\)
The favourable area is divided by the total area, so the probability cannot exceed one

Difficulty levels

challengeCalculate an exact area ratio after identifying a nested, clearance or polygon event region

Questions about this topic

Why is area used instead of perimeter?+

The point or centre is uniform over a two-dimensional region, so equal likelihood is measured by area. Perimeter measures only the boundary

Why must an object clearance region be reduced?+

The centre must stay far enough from each edge for the whole object to remain on the board. Subtract the object width and height before finding the allowable area

Can a geometric probability be greater than one?+

No. The favourable region is contained in the total allowable region, so its area ratio is from zero to one

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 8: Probability and counting

In this chapter

Finite events, probability assignments and complementsRepresent a multi-stage finite sample spaceCompound events, the addition rule and mutual exclusivityComplete and use a two-way probability tableConditional probability from exact representationsDistribution of a finite random variableDesign and interpret a probability simulationCount restricted selections and calculate exact probabilitiesCount arrangements with product, sum and restriction rulesTest exact independenceExact probability of repeated independent successesMulti-stage probability, multiplication and total probabilityProbability from a uniform area modelProbabilities of sequences of independent events