If points A and B are randomly placed on the circumference

There is one little point you missed (using the method you have used to find the way the chord is chosen). Say you put the first point anywhere on the circumference. Now, you have found that if you put the other point on 1/6th of the circumference (right next to the first point), the chord length will be less than or equal to 2. But you have to consider the 1/6th of the circle on the other side of the point too. Say, in your diagram, the left vertex of the triangle lying on the circle is A, the first point. Now B can be to the right of A or to the left of A. So you can put B on 1/3rd of the circle and still get a chord less than or equal to 2. So answer will be 2/3.
_________________

Not a GMAT question. Looks like: [Bertrand paradox]

One of the solutions:

Let C be the center of the circle.

Place point A on any place on the circumference. Now, if \(\angle{ACB}=60\), then chord \(AB=raius=2\), so if \(\angle{ACB}<60\), \(AB<raius=2\). But point B can be placed either clockwise or anticlockwise from A. Which means that if B falls in 60 degrees in either of direction from A, AB will be less than radius. Total circumference 360 degrees restricted area 60*2=120 degrees so \(P(AB>2)=\frac{360-120}{360}=\frac{2}{3}\).

But answer can also be \(\frac{3}{4}\) or \(\frac{\sqrt{3}}{2}\).

Don't worry about this question.
_________________

I don't know whether I am correct or not.

Probability of chord greater than 2 = 1- probability of chord less than or equal to 2.
Probability of chord less than or equal to 2 = (Length of circumference of shaded circle)/ Length of circumference = [(60/360)2*pi*2] / [2*pi*2] ie. [(4/6)*pi]/[4*pi] ie 1/6.
Hence, Probability of chord greater than 2 = 1-1/6 =5/6. Choice E.

Merging similar topics. Please refer to the solutions above.
_________________

IMO D - 2/3

Probability = required length of arc/ circumference

Take chord AB and join A and B with the center.
Find the angle subtended by the chord AB at the center

It will be 60 degrees as sin(AOD) = 1/2 => angle AOD = 30
=> angle AOB = 60
Where D is mid point of the chord

Now for the AB > 2 the angle subtended by the chord will be > 60 on one half of the circle and 60 degree on other half of the circle.
Total angle = 120 = 2π/3
So total excluded arc = angle * radius = 2πr/3
So required arc = 2πr - 2πr/3 = 4πr/3

Probability = (4πr/3) / ( 2πr) = 2/3
hence D

how 3/4 is possible?

Bunnel, could you analyze this question...
how-many-times-swimmers-meet-93953.html#p722693

Is this GMAT question? The user has not posted the OA and OE, though I have solved it but need expert comments.

https://mathforum.org/library/drmath/view/66793.html
https://mathworld.wolfram.com/BertrandsProblem.html

Again, please don't worry about this question.



Again not a GMAT question but 13 seems to be a correct answer.
_________________

This is the way I did it:

I drew a circle and drew a line from the center to the point, forming the radius. I asked myself, if I were to pick a another point on the circle for which the chord is equal to 2, what would the angle be of the radius to the first point and the second point? Well the radius of the circle is 2, so if I wanted to make a chord of length two, all I would have to do it draw a 60,60,60, equaliteral triangle from the center to the first point and two the first point making a 60 degree angle between them. The same long is used again, If I were to draw a third point, in the other direction to form a chord of length two, the angle between the center and the first and third point would be 60*. Hence for any single point on the circle, the two point that can be possible be drawn to generate a length of 2, from that first point span 60+60=120*. Everything above that degree and we have a chord bigger than 2. Now 1-(120/360*2*pi*r)=1/3. Hope that made sense

2/3. There are only 2 chords of 2 in that can be drawn..on adjacent equilateral triangles. These triangles take up 120 degrees of the circle...leaving 240 degrees outside the probable area...hence 240/360=2/3.

Thanks Karishma!!!. I have not thought of that possibility. So I have to add another 1/6.

I think one way of answering this question is:

Let us assume that the center is A and the two ends of the chord are B and C.

Let us first assume that the length of the chord is 2.

If the length of the chord has to be 2 to start with. The triangle created by drawing lines from the two ends of the chord to the center would be an equilateral triangle.

Which means angle BAC would be 60 degrees. If the angle BAC is less than 60 then the length of the chord would be less than 2 and if it is more than 60 it would be greater than 2.

This means that there are 120 possibilities for angle BAC where the length of BC would be greater than 2.

The probability would therefore be 120/180 = > 2/3

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

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 either side of A is out of bounds.
i.e 120 degress of the circumference is out of bounds.
So probability = 360-120/360 = 240/360 = 2/3

Answer is D

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.


_________________