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

A point in the figure selected at random. find the probability that the point will-example-1
User Lost Baby
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1 Answer

4 votes

Answer:

The probability that the point will be in the part that is NOT shaded is about 20% ⇒ 4th answer

Explanation:

* Lets look to the figure

- There are four circles inscribed in a square

- The four circles touched each other and touched the four

sides of the square

∴ The side of the square = twice the diameter of a circle

- If the side of the square is l and the diameter of the circle is d

∴ l = 2d ⇒ divide the two sides by 2

∴ d = (1/2) l

∵ The radius of the circle = (1/2) the diameter

∴ r = (1/2) d

∵ d = (1/2) l

∴ r = (1/2)(1/2) l = (1/4) l

* Now lets find the area that NOT shaded

∵ The area of the square = side × side

∴ The area of the square = l × l = l²

∵ The area of the circle = πr²

∵ r = (1/4) l

∴ r² = [(1/4) l]² = (1/4)² × l² = (1/16) l²

∴ The area of one circle = (1/16)πl²

- The shaded part is the four circles

∴ The shaded area = 4 × (1/16)πl² = (1/4)πl²

- The part is not shaded = Area of the square - Area of the shaded part

∴ The area of not shaded = l² - (1/4)πl² ⇒ take l² as a common factor

∴ The area of not shaded = l²(1 -1/4 π)

- The probability that the point will be in the part that not shaded is

area of the part not shaded/area of the square

∴ P = l²(1 - 1/4 π)/l² ⇒ cancel l² from up and down

∴ P = (1 - 1/4 π )/1 = 0.2146 ≅ 0.2

- Chang it to percent number

∴ P = 0.2 × 100% = 20%

* The probability that the point will be in the part that is NOT shaded

is about 20%

∴ r = (1/2) d

User Aleksei Tirman
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