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505-555 Level|   Geometry|   Probability|      
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Bunuel
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In the accompanying diagram, a fair spinner is placed at the center O 02 Dec 2021, 09:00
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83% (01:24) correct 17% (01:32) wrong based on 86 sessions

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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 labeled X, Y and Z. If ∠BOC = 45 degrees. What is the probability that the spinner will land in the region X?

A) 1/8
B) 3/8
C) 1/2
D) 1/4
E) None of the Above


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Sushen
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In the accompanying diagram, a fair spinner is placed at the center O 02 Dec 2021, 18:31
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45 = 1/8 (360)
x + y + z = 1 (whole circle)
y = 180 = 1/2 (360)

x + y + z = 1 (whole circle.)
x + 1/2 + 1/8 = 1
x = 3/8

B is the answer
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In the accompanying diagram, a fair spinner is placed at the center O 30 Dec 2021, 08:08
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Area of a sector = (πr^2θ)/360
Area of Z = (πr^2*45)/360 = πr^2/8
Area of Y = πr^2/2
Area of X = Area of Circle - Area of Z - Area of Y
Area of X = πr^2 - πr^2/8 - πr^2/2
Area of X = 3/8πr^2

Probability = Favorable Area/Total Area = 3/8πr^2/πr^2 = 3/8 (B)
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In the accompanying diagram, a fair spinner is placed at the center O 02 Jan 2023, 09:00
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[quote="Bunuel"]
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 labeled X, Y and Z. If ∠BOC = 45 degrees. What is the probability that the spinner will land in the region X?

A) 1/8
B) 3/8
C) 1/2
D) 1/4
E) None of the Above


Area of an arc = pi*r^2*(theta/360)
Area of region X = pi*r^2*(135/360)

Probability = pi*r^2*(135/360)/pi*r^2 ==> 135/360 ==> 15/40 ==> 3/8
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In the accompanying diagram, a fair spinner is placed at the center O 02 Jan 2023, 23:50
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Bunuel

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 labeled X, Y and Z. If ∠BOC = 45 degrees. What is the probability that the spinner will land in the region X?

A) 1/8
B) 3/8
C) 1/2
D) 1/4
E) None of the Above

Solution:

  • Angle subtended at center for region X \(=\angle AOC=180-45=135^o\)
  • Tjis \(135^o\) happens to be the favorable case whereas the whole circle \(360^o\) happens to be the total case
  • So, the probability of the spinner landing in region \(X =\frac{135}{360}=\frac{3}{8}\)\(\)

Hence the right answer is Option B
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