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Re: probability q'n [#permalink]
Let us call O, AB, R the center, the chord, and the radius of the circle, respectively.
\(AB\geq2\sqrt{3}\) and \(R=2\)

It is remarked that when \(AB\geq2\sqrt{3}\), the corresponding central angle \(AOB\geq120^{\circle}\)

Result = \(\frac{120^{\circle}}{360^{\circle}}=\frac{1}{3}\)
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Re: probability q'n [#permalink]
sandeep800 wrote:
If points A and B are randomly selected from the circumference of a circle with diameter 4, what is the probability that the length of chord AB will be at least 2sqrt{3}?

can anyone please solve?


Corresponding central angle for a dimater is 180, so let x be a corresponding central angle for a chord of 2sqrt3

So, x/180=2sqrt3/4

Estimate x
Is this approach correct?
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Re: probability q'n [#permalink]
I think we can use Integral (Area under the curve) for this problem. Surely, not within scope of GMAT.
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Re: probability q'n [#permalink]
bogos wrote:
Let us call O, AB, R the center, the chord, and the radius of the circle, respectively.
\(AB\geq2\sqrt{3}\) and \(R=2\)

It is remarked that when \(AB\geq2\sqrt{3}\), the corresponding central angle \(AOB\geq120^{\circle}\)

Result = \(\frac{120^{\circle}}{360^{\circle}}=\frac{1}{3}\)


How did you arrive at the 120 deg? My thinking is that the distance to a chord of length 2sqrt(3) is 1 from the center and the radius is 2, should then the probability not be 50%?



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