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 $1.5.$

blackpanther Feb 14, 2024

#1**+2 **

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

#1**+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