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Two points on a circle of radius $1$ are chosen at random.  Find the probability that the distance between the two points is at most $1.5.$

 Feb 14, 2024

Best Answer 

 #1
avatar+394 
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Suppose point A is the first point choosen. To satisfy the contition, the second point has to be on the major arc of BB'\(\). But the question now is how do we find the value of \(\angle{BO{B}^{'}}\)? Well, we know \(\angle{BO{B}^{'}}=\angle{AO{B}^{'}}*2\). There are many ways to find angle AOB', but I choose to use law of cosines. I won't get into the details, but

 \(\cos(\angle{AO{B}^{'}})=\frac{1+1-\frac{9}{4}}{2}=-\frac{1}{8}\), so \(\angle{AO{B}^{'}}=\arccos(-\frac{1}{8})\), and \(\angle{BO{B}^{'}}=2*\arccos(-\frac{1}{8})\)\(\).

Supposing you are calculating in radians, the probability is \(\frac{2*\arccos(-\frac{1}{8})}{2\pi}\), or \(\frac{\arccos(-\frac{1}{8})}{\pi}\), which is approximately 0.53989, or 53.989%.

 Feb 15, 2024
 #1
avatar+394 
+2
Best Answer

Suppose point A is the first point choosen. To satisfy the contition, the second point has to be on the major arc of BB'\(\). But the question now is how do we find the value of \(\angle{BO{B}^{'}}\)? Well, we know \(\angle{BO{B}^{'}}=\angle{AO{B}^{'}}*2\). There are many ways to find angle AOB', but I choose to use law of cosines. I won't get into the details, but

 \(\cos(\angle{AO{B}^{'}})=\frac{1+1-\frac{9}{4}}{2}=-\frac{1}{8}\), so \(\angle{AO{B}^{'}}=\arccos(-\frac{1}{8})\), and \(\angle{BO{B}^{'}}=2*\arccos(-\frac{1}{8})\)\(\).

Supposing you are calculating in radians, the probability is \(\frac{2*\arccos(-\frac{1}{8})}{2\pi}\), or \(\frac{\arccos(-\frac{1}{8})}{\pi}\), which is approximately 0.53989, or 53.989%.

hairyberry Feb 15, 2024

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