Find the probability that a point chosen at random will lie in the shaded area

A. 0.32
B. 0.42
C. 0.02
D. 0.72

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konrad509 konrad509 · Answer 1 · 2018-07-10T17:57:06+02:00

[tex]|\Omega|=\pi \cdot9^2=81\pi\\ |A|=\dfrac{1}{2}\cdot18\cdot9=81\\\\ P(A)=\dfrac{81}{81\pi}=\dfrac{1}{\pi}\approx0.32\Rightarrow \text{A}[/tex]

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setukumar345 setukumar345 · Answer 2 · 2022-06-03T20:38:11+02:00

The probability that a point is chosen at random will lie in the shaded area is 0.32.

Here favorable outcomes will be the shaded portion and the total outcome will be the total circular portion.

Favorable Outcome = Area of shaded portion = [tex]\frac{1}{2} (18)(\frac{18}{2}) = 81 \ inch^2[/tex]
Total Outcome = Total area = [tex]\pi \ ( \frac{18}{2} )^{2} = 81 \ \pi \ inch^{2}[/tex]
∴ Probability, P(E) = [tex]\frac{81}{81\pi } =\frac{1}{\pi } = 0.32[/tex]

Thus, the probability that a point chosen at random will lie in the shaded area is 0.32.

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