In the accompanying diagram a fair spinner is placed at centre O of circle. Diameter AOB and radius . . .

Question : In the accompanying diagram a fair spinner is placed at centre O of 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 spinner will land in the region X?

Doubt by Vanshika

Solution : 

In this question, we will going to use the concept of area of a sector (part) of a circle.
Area of sector = (θ/360°)×πr²

Total possible area = area of circle = πr²

AOB is a diameter.

∴ ∠AOC+∠BOC = 180° (Linear Pair)

∠AOC+45° = 180°
∠AOC = 180°-45°
∠AOC = 135°

Favorable area = Area of sector ACC = (135°/360°)×πr²

We know,

P(E) = (No. of favorable outcomes) / (Total No. of possible outcomes)

P(Spinner will land in the region X)

= [(135°/360°)×πr²]/πr²

=135°/360°

= 27/72

= 3/8