Probability and Area

Probability can also relate to the areas of geometric shapes. The following are some examples of probability problems that involve areas of geometric shapes.

Example:

A dart is thrown at random onto a board that has the shape of a circle as shown below. Calculate the probability that the dart will hit the shaded region. (Use π = 3.142)

Solution:

Total area of board = 3.142 × 14 2 = 615.83 cm²

Area of non-shaded circle = 3.142 × 7 2 = 153.99 cm²

Area of shaded region = 615.83 – 153.99 = 461.84 cm² = 462 cm² (rounded to whole number)

Probability of hitting the shaded region:

=

Example:

The figure shows a circle divided into sectors of different colours.

If a point is selected at random in the circle, calculate the probability that it lies:

a) in the red sector

b) in the green sector.

c) in any sector except the green sector.

Solution:

a) Area of red sector = × area of circle

Probability that the point lies on red sector =

b) Area of green sector = × area of circle

Probability that the point lies on green sector =

c) in any sector except the green sector.

Probability that the point does not lie in the green sector =

Example:

In the figure below, PQRS is a rectangle, and A, B, C, D are the midpoints of the respective sides as shown.

An arrow is shot at random onto the rectangle PQRS. Calculate the probability that the arrow strikes:

a) triangle AQB.
b) a shaded region.
c) either triangle BRC or the unshaded region.

Solution:

a) Let PQ = 2x and QR = 2y. Then, AQ = x and QB = y.

Area of rectangle PQRS = 2x × 2y = 4xy

Area AQB = xy

Probability of striking triangle AQB = xy ÷ 4xy =

b) All the shaded triangles are equal.

Total area of shaded regions = 4 × xy = 2xy

Probability of striking a shaded region = 2xy ÷ 4xy =

c) Area of unshaded region = 4xy – 2xy = 2xy

Probability of striking unshaded region = 2xy ÷ 4xy =

Area of triangle BRC = xy

Probability of striking triangle BRC=

Probability of striking triangle BRC or unshaded region =

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