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Question
In the accompanying diagram a fair spinner is placed at the center O of the circle.
Diameter AOB and radius OC divide the circle into three regions labelled X, Y and Z. If
∠BOC = 45°. What is the probability that the spinner will land in the region X?(See fig)
Solution
Given ∠BOC = 45°
∠AOC = 180 – 45 = 135°
Area of circle = 𝜋r2
Area of region × =`theta/(360^@)`× 𝜋𝑟2
= `135/360× pir^2 = 3/8pir^2`
Probability that the spinner will land in the region
X = `"Area of region x"/"total area of circle"` =`( 3/8pir^2)/(pir^2) = 3/8`
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