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Circle question help

This topic has 3 expert replies and 1 member reply

Circle question help

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If points A and B are randomly placed on the circumference of a circle with radius 2, what is the probability that the length of chord AB is greater than 2?
A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

I keep getting 5/6....
I thought I can draw two triangles (30'60'90' special triangle) with base length 1 each and thus add up to 2, and the two angles add up to 60 degrees. so I got 300/360, but the answer is different from what I got.

Can someone please provide explanation?

Thanks much in advance

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Quote:
If points A and B are randomly placed on the circumference of a circle with radius 2, what is the probability that the length of chord AB is greater than 2?
A)1/4
B)1/3
C)1/2
D)2/3
E)3/4
Question rephrased: Once A has been placed on the circle, what is the probability that B can be placed on the circle such that AB>2?



AB=2 if B is placed to the left or right of A as shown by B₁ and B₂ in the figure above.
If either B₁ or B₂ moves CLOSER to A, then AB<2.
Implication:
Since ∠B₁OB₂ = 60+60 = 120º, and 120/360 = 1/3, AB≤2 for 1/3 of the circle.
Thus, AB>2 for the remaining 2/3 of the circle.
Resulting probability:
P(AB>2) = 2/3.

The correct answer is D.

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This question is a version of a much older problem -- that of choosing a random chord on a circle -- that actually doesn't have a single answer, at least without a clearer explication of the process by which the chord is chosen. (The way this particular version is posed is better, but not perfect.)

Here's Mosteller's explanation of the problem from his Fifty Challenging Problems in Probability:

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jchung402 wrote:
If points A and B are randomly placed on the circumference of a circle with radius 2, what is the probability that the length of chord AB is greater than 2?
A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4
Thanks much in advance
If the length of the chord is 2, then the radii joining the ends of the chord to the centre form an equilateral triangle. i.e the angle between A & B from the centre has to be 60.

If A is a random point on the circumference then B can be any point further than 60 degrees of A on either side of A.

i.e 60 degrees on Both the sides of A is out of the bounds.

i.e a Total of 120 degrees of the circumference is out of bounds.

So probability = (360-120)/360 = 240/360 = 2/3

Answer: Option D

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jchung402 wrote:
If points A and B are randomly placed on the circumference of a circle with radius 2, what is the probability that the length of chord AB is greater than 2?
A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4
We'll begin by arbitrarily placing point A somewhere on the circumference.




So, we want to know the probability that a randomly-placed point B will yield a chord AB that is at least 2 cm long.
So, let's first find a location for point B that creates a chord that is EXACTLY 2 cm long.




There's also ANOTHER location for point B that creates another chord that is EXACTLY 2 cm long.




IMPORTANT: For chord AB to be greater than or equal to 2 cm, point B must be placed somewhere along the red portion of the circle's circumference.



So, the question really boils down to, "What is the probability that point B is randomly placed somewhere on the red line?"
To determine this probability, notice that the 2 cm chords are the same length as the circle's radius (2 cm)



Since these 2 triangles have sides of equal length, they are equilateral triangles, which means each interior angle is 60 degrees.



The 2 central angles (from the equilateral triangles) add to 120 degrees.
This means the remaining central angle must be 240 degrees.


This tells us that the red portion of the circle represents 240/360 of the entire circle.
So, P(point B is randomly placed somewhere on the red line) = 240/360 = 2/3

Answer: D

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