If two points, A and B, are randomly placed on the circumference

Question

If two points, A and B, are randomly placed on the circumference of a circle with circumference 12pi inches, what is the probability that the length of chord AB is at least 6 inches?

(A) 1/(2pi) 
(B) 1/pi 
(C) 1/3
(D) 2/pi
(E) 2/3

*kudos for all correct solutions

BrentGMATPrepNow · Accepted Answer

pitroncoso's solution is perfect. 
Below is a very similar solution (with a few diagrams to help students visualize the solution)

Let's first determine the details of this circle. 
For any circle, circumference = (diameter)(pi)

The circumference of the given circle is 12pi inches, so we can write: 12pi inches= (diameter)(pi)
This tells us that the diameter of the circle = 12 inches
It also tells us that  the radius of the circle = 6 inches

Okay, now let's solve the question. 
We'll begin by arbitrarily placing point A somewhere on the circumference. 
 https://i1168.photobucket.com/albums/r500/GMATPrepNow/c1_zpsauzqfzb9.png

So, we want to know the probability that a randomly-placed point B will yield a chord AB that is at least 6 inches long.
So, let's first find a location for point B that creates a chord that is EXACTLY 6 inches. 
 https://i1168.photobucket.com/albums/r500/GMATPrepNow/c2_zpsjtall5xw.png

There's also another location for point B that creates another chord that is EXACTLY 6 inches. 
 https://i1168.photobucket.com/albums/r500/GMATPrepNow/c3_zpsbdyvle98.png

IMPORTANT: For chord AB to be  greater than or equal to 6 inches , point B must be placed somewhere along the red portion of the circle's circumference. 
 https://i1168.photobucket.com/albums/r500/GMATPrepNow/c4_zpsovzjbvhw.png

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 6-inch chords are the same length as the circle's radius (6 inches)
 https://i1168.photobucket.com/albums/r500/GMATPrepNow/c5_zpspqjnr3d6.png

Since these 2 triangles have sides of equal length, they are equilateral triangles, which means each interior angle is 60 degrees. 
 https://i1168.photobucket.com/albums/r500/GMATPrepNow/c6_zpssxbrteja.png

The 2 central angles (from the equilateral triangles) add to 120 degrees. 
This means the remaining central angle must be 240 degrees. 
 https://i1168.photobucket.com/albums/r500/GMATPrepNow/c7_zpshzmhhj3r.png

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: E

pitroncoso · Answer

(Another method)

Raduis is 6 (2pi*r=12pi=2pi*6)
Fix point A in the circumference (let's choose the north pole)

From A, draw the two possible chords of length 6 ('left' and 'right')
Ending points are X and Y.
Join A, X and Y to the center O.

We have two equilateral triangles (each side is 6), so angle XOY is 120 degrees.

Now, point B of our 'chord grater than 6' can be chose anywhere outside arch YX. As inside is 120, outside is 240 degrees.

So the probability is 240 (degrees)/ 360 (degrees) = 2/3 letter E

Jayeshvekariya · Answer

Tough call...

Tried and failed to do...need to brush up my skills

beefcakeharsh · Answer

Probability questions involving circles are best solved using area or angle subtended on the centre.

As 12pi is the area, diameter of circle is is 12 in. Taking the limiting case of 6 in chord - A chord of 6 inch will subtend 60 degrees on the the centre. (Equilateral Triangle with the radius)

Any chord more than 6 inch will subtend more than 60 degrees on the centre.

So the Probability of the chord being in that sector is 240/360 = 2/3

Answer is E

Zoser · Answer

Can you please explain why you considered two points for B not just one? is not the answer 300/360 instead?

BrentGMATPrepNow · Answer

You bet. 
Once point A is placed on the circumference, there are TWO chords of length 6 that can be connected to point A: one chord is to the left of point A, and the other chord is to the right of point A. 
If we don't consider both points then we will be including a portion of the circle where the chord AB can be less than 6.

Does that help?

Cheers,
Brent

Zoser · Answer

Common sense. Thanks a lot

sahilvijay · Answer

Quite simple ques
see the figure attached
2pi R = 12 pi => R=6
when AB is joined ( length = 6 minimum ) consider AB as 6

the angle made by chord at center of the circle will be 60 degrees

Now consider same chord on opposite side
same angle will be 60

so the angle 60 will be the angle on which any 2 points taken beyond AB will be less than 6

There fore 2 sides were there => 1 - 120/360 = 2/3

THIS WAS 1 METHOD

2ND METHOD:

see the picture attached
between 2 chords of 6 in length
120 degree is the angle on 1 side on which any point taken suppose A and on the other side if any point taken between that 120 degree on the circle will make >= 6 chords' length.

therefore P(total) = 120+120 / 360 = 240/360 = 2/3

sahilvijay · Answer

Hello Zoser you can consider 2 parallel chords as explained in my answer. And find the extremities of 6 then calculate the angle, 
i guess it is self explanatory.

bumpbot · Answer

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If two points, A and B, are randomly placed on the circumference

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If two points, A and B, are randomly placed on the circumference

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