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(D) Let the two points be $A$ and $B$ on the circumference of a circle with center $O$ and radius $r$.Let $	heta$ be the central angle $\angle AOB$ s

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(D) Let the two points be $A$ and $B$ on the circumference of a circle with center $O$ and radius $r$.Let $	heta$ be the central angle $\angle AOB$ subtended by the chord $AB$ at the center.The length of the chord $AB$ is given by $2r \sin(\frac{	heta}{2})$.We want the distance $AB \ge r$,which implies $2r \sin(\frac{	heta}{2}) \ge r$.This simplifies to $\sin(\frac{	heta}{2}) \ge \frac{1}{2}$,which means $\frac{	heta}{2} \ge 30^{\circ}$ or $	heta \ge 60^{\circ}$.Since the points are chosen randomly,the angle $	heta$ can range from $0$ to $180^{\circ}$ (as the distance is symmetric for $	heta$ and $360^{\circ}-	heta$).The total range of the angle $	heta$ is $180^{\circ}$.The favorable range for $	heta$ is $60^{\circ} \le 	heta \le 180^{\circ}$.The probability is the ratio of the favorable range to the total range: $P = \frac{180^{\circ} - 60^{\circ}}{180^{\circ}} = \frac{120^{\circ}}{180^{\circ}} = \frac{2}{3}$.

Two points are randomly chosen on the circumference of a circle of radius $r$. The probability that the distance between the two points is at least $r$ is equal to

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