A) point m is chosen random inside the big circle the radii of the small, middle and big circles are 4 cm, 7 cm, 10 cm respectively. what is the probability that point m will be in part a? b) point m is chosen randomly inside the big circle the radii of the small, middle and big circles are 4 cm, 7 cm, 10 cm respectively. what is the probability that point m will be in part b? c) point m is chosen randomly inside the big circle the radii of the small, middle and big circles are 4 cm, 7 cm, 10 cm respectively. what is the probability that point m will be in part c?

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Area of circle = πr²

Question 1st)

The probability that the a point chosen at random will be in part A will be equal to = (Area of Part A) / (Total Area)

Area of Part A = Area of Big Circle - Area of Middle Circle

So, Area of Part A = 100π - 49π = 51π

Total Area = Area of Big Circle = 100π

Thus, the probability that the point is in Part A =

Question 2nd)

The probability that the a point chosen at random will be in part B will be equal to = (Area of Part B) / (Total Area)

Area of Part B = Area of Middle Circle - Area of Smaller Circle

So, Area of Part B = 49π - 16π = 33 π

Total Area = Area of Big Circle = 100π

Thus, the probability that the point is in Part B =

Question 3rd:

The probability that the a point chosen at random will be in part C will be equal to = (Area of Part C) / (Total Area)

Area of Part C = Area of Small Circle

So, Area of Part C = 16π

Total Area = Area of Big Circle = 100π

Thus, the probability that the point is in Part C =

6 equilateral triangles