Question

A point in the figure is selected at random. Find the probability that the point will be in the part that is NOT shaded

Answer 1

Let r be the radius

Area of the shaded part (which is 2 complete circles)

= 2 x πr²

= 2πr²

Area of the unshaded part (which is a square that has a length 2r)

= 2r x 2r

= 4r²

Probability that the point is in the unshaded area

`\frac{\text{area of square}}{\text{area of square + 2 circles}}`

`= (4\pi r^(2))/(2\pi r^(2) + 4\pi r^(2)) = (2r^(2)(2))/(2r^(2)(\pi + 2)) = (2)/(\pi + 2) = 0.389`

Answer: 0.4 (nearest tenth)