# In the accompanying diagram a fair spinner is placed at the centre O of the circle Diameter

Question:

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 XY and Z. It ∠BOC = 45°. What is the probability that the spinner will land in the region X? (in the given figure).

Solution:

Given: A fair spinner is placed at the centre O of the circle. Diameter AOB and radius OC divide the circle into three regions labeled X, Y and Z and angle

To find: Probability that the spinner will land in X region?

Total angle of circle is 360°.

$\angle \mathrm{AOC}+\angle \mathrm{BOC}=180^{\circ}$ (Straight angle)

$\angle \mathrm{AOC}+45^{\circ}=180^{\circ}$

$\angle \mathrm{AOC}=180^{\circ}-45^{\circ}$

$\angle \mathrm{AOC}=135^{\circ}$..............(1)

We know that PROBABILITY $=\frac{\text { Number of favourable event }}{\text { Total number of event }}$

Hence probability of "spinner will land in $X$ region" is $\frac{135}{360}=\frac{3}{8}$