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Probability

by yellowho » Mon Feb 14, 2011 12:29 am
If two points A and B are randomly placed on the circumference of a circle with a diameter of 4
inches, what is the probability that the length of chord AB is at least 2 inches?

My method:

radius= 2
Circumference= 4pi
4pi(k/360)=2 since we want to know what % of 360 will create 2. anything higher than that will create a chord higher than 2.

k= 180/pi

estimate roughly 60 degree. If you split a circle up into 360 points. 120 of them will be 2 or less. roughly 2/3 will more.

To me this feels very inaccurate especially when it didn't say "Approximate." I'm sure there's a better way. Please enlighten me.
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by Geva@EconomistGMAT » Mon Feb 14, 2011 1:57 am
yellowho wrote:If two points A and B are randomly placed on the circumference of a circle with a diameter of 4
inches, what is the probability that the length of chord AB is at least 2 inches?

My method:

radius= 2
Circumference= 4pi
4pi(k/360)=2 since we want to know what % of 360 will create 2. anything higher than that will create a chord higher than 2.

k= 180/pi

estimate roughly 60 degree. If you split a circle up into 360 points. 120 of them will be 2 or less. roughly 2/3 will more.

To me this feels very inaccurate especially when it didn't say "Approximate." I'm sure there's a better way. Please enlighten me.
At least questions are best solved using 1-p(unwated outcome) - in this case, 1- P(distance of 2 or lower)
Let's find the forbidden scenario: probability of distance of 2 or lower. Split the process into First point, 2nd point, and find the probability of a the outcome for each point.
For the first point, we don't really care where it is placed on the circle: the probability of a wanted outcome is 1.
for the 2nd point, we want the probability of a distance of 2 from the first. If the chord AB is exactly 2, then the result is an equilateral triangle with the two radii of 2, so the central angle is degrees. That's the turning point: if the central angle which defines the arc AB is 60 degrees or lower, the result is a chord of length 2 or smaller.
Note however, that the angle AB could be in either direction: if we placed A somewhere in the circle, B can go 60 degrees to the clockwise, or counter-clockwise, so the total angle that will create a chord of 2 or less is 120 degrees.

thus, the p(2 or less) is 120/360 - meaning that the probability of at least 2 is 1-120/360 = 240/360 = 2/3.
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by garuhape » Mon Feb 14, 2011 4:50 am
@ Geva,

where did you get the 60° from? Even if I calculate it accurately with my calculator I get 57....°

My steps are:

circumference: 4phi
distance between A and B: 2 to both sides: 4

Therefore: 4phi-4=12.56-4=8.56 [the rest of the circumference]

Then: 8.56/12.56 = 856*10/1256= 6.8/10=0.68=68%

MY result is similar but different and more complicated with the last division. Pls Help
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by Anurag@Gurome » Mon Feb 14, 2011 5:52 am
garuhape wrote:My steps are:

circumference: 4phi
distance between A and B: 2 to both sides: 4

Therefore: 4phi-4=12.56-4=8.56 [the rest of the circumference]
The problems says 2 is the chord length.
But you're assuming it as the arc length.
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by Geva@EconomistGMAT » Mon Feb 14, 2011 6:57 am
garuhape wrote:@ Geva,

where did you get the 60° from? Even if I calculate it accurately with my calculator I get 57....°

My steps are:

circumference: 4phi
distance between A and B: 2 to both sides: 4

Therefore: 4phi-4=12.56-4=8.56 [the rest of the circumference]

Then: 8.56/12.56 = 856*10/1256= 6.8/10=0.68=68%

MY result is similar but different and more complicated with the last division. Pls Help
the 60 stems from the fact that the triangle created by chord AB, and the two radii from the center to A and B is equilateral when the chord is equal to 2: an equilateral triangle has all three angles at 60 degrees.
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by nubu » Mon Feb 14, 2011 7:32 am
Any time we place 2 points, one must come first. Let's say A.
So, now we just care about the probability that B is placed to make AB at least 2.
when Chord AB = 2, we have AB=OA=OB=2, so AOB = 60 degree (O is a center). The point is that we can place B at the RIGHT SIDE or the LEFT SIDE of A so "probability that AB is at least 2" = 1 - 2x(60/360) =2/3

Hope this help.
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by garuhape » Mon Feb 14, 2011 11:48 am
Thanks, that was very helpful.
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