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In the accompanying diagram a fair spinner is placed at the centre O of the circle. Diameter AOB and radius OC divide the circle into three regions labelled X, Y and Z. If $\angle BOC = 45^o$. What is the probability that the spinner will land in the region X?
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Given:
A fair spinner is placed at the centre O of the circle.
Diameter AOB and radius OC divide the circle into three regions labelled X, Y and Z.
$\angle BOC = 45^o$.
To do:
We have to find the probability that the spinner will land in the region X.
Solution:
Let $r$ be the radius of the circle.
$\angle \mathrm{BOC}=45^{\circ}$
$\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ are three sectors of the circle with the central angle $180^{\circ}-45^{\circ}=135^{\circ}, 180^{\circ}$ and $45^{\circ}$ respectively.
Therefore,
Probability that the spinner will land in the region $X=\frac{\text { Area of sector } \mathrm{X}}{\text { Area of circle }}$
$=\frac{\pi r^{2} \times \frac{135^{\circ}}{360^{\circ}}}{\pi r^{2} \times \frac{360^{\circ}}{360^{\circ}}}$
$=\frac{135^{\circ}}{360^{\circ}}$
$=\frac{3}{8}$
The probability that the spinner will land in the region $X$ is $\frac{3}{8}$.
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