Question

Find the probability that a point chosen at random would lie in the shaded area of the figure. Round to the nearest tenth of a percent.

with the steps and explanation

Answer 1

Explanation:

The required probability is simply the (shaded area)/(area of the whole box)

The area of the whole box would be:

6r*2r = 12 r^2

The shaded area is tough to calculate, but you can get the

white area instead.

The left hole (white area) consists of a semi-circle and a rectangle.

The center hole is a circle.

The right hole (white area) consists of a semi-circle and a rectangle.

The 2 semi-circles in the left and right hole combine to be a circle.

The 2 rectangles in the left and right hole combine to be a square.

The total area of holes are:

(area of circle)*2 + (area of square) ### 2 circles 1 square in total

=r^2*pi*2 + (2r)^2

=r^2*pi*2 +4*r^2

then we can get the shaded area by:

(area of the whole box) - (The total area of holes)

= 12 r^2 - (r^2*pi*2 +4*r^2)

=8*r^2 - r^2*pi*2

Finally:

(shaded area)/(area of the whole box)

=(8*r^2 - r^2*pi*2)/(12 r^2)

=(8-pi*2)/(12)

=0.143067891

=14.3% (corr. to 3 sig. fig.)