Question

What is the probability that the inclosed figure formed by joining three points choosen randomly on the circumference of the circle will contain the center of the circle?​

Answer 1

Answer:

Solution:

$\sf{If \ all \ points \ lie \ on \ circumference \ of \ the}$

$\sf{circumference \ of \ circle.}$

$\sf{Let \ points \ be \ p_{1}, \ p_{2} \ and \ p_{3}}$

$\sf{and \ centre \ be \ C.}$

$\sf{If \ initial \ ray \ is \ formed \ by \ joining \ centre}$

$\sf{and \ p_{1} \ and \ terminal \ ray \ by \ joining}$

$\sf{centre \ and \ p_{3}}$

$\sf{[Here \ p_{1} \ and \ p_{3} \ are \ vertices \ of}$

$\sf{triangle \ where \ p_{2} \ will \ always \ lie \ on \ the}$

$\sf{arc \ formed \ by \ initial \ ray \ and \ terminal \ ray.]}$

$\sf{Let's \ measure \ angle \ between \ initial \ ray}$

$\sf{and \ terminal \ ray \ in \ anti \ clockwise \ direction.}$

$\sf{If \ the \ angle \ between \ initial \ ray \ and \ terminal}$

$\sf{ray \ is \ less \ than \ 180^\circ \ then \ after \ joining}$

$\sf{p_{1} \ and \ p_{3} \ the \ centre \ will \ be \ not}$

$\sf{in \ closed \ in \ the \ triangle.}$

$\sf{In \ same \ way, \ if \ angle \ between \ initial}$

$\sf{ray \ and \ terminal \ ray \ is \ more \ than \ 180^\circ}$

$\sf{centre \ will \ be \ in \ closed \ in \ the \ triangle. }$

$\sf{Also,}$

$\sf{If \ measure \ of \ angle \ between \ terminal}$

$\sf{ray \ and \ initial \ ray \ is \ 180^\circ \ the \ center}$

$\sf{will \ lie \ on \ the \ on \ of \ the \ side.}$

$\sf{Hence, \ Total \ number \ of \ outcome \ will \ be \ 2n}$

$\sf{(approx) \ if \ n \ is \ number \ of \ outcomes \ for}$

$\sf{condition \ if \ centre \ is \ in \ closed \ in}$

$\sf{the \ triangle \ formed \ by \ joining \ points.}$

$\sf{Probability=\dfrac{Number \ of \ outcomes}{Total \ number \ of \ outcomes}}$

$\sf{\therefore{Probability=\dfrac{n}{2n}}}$

$\sf{\therefore{Probability=\dfrac{1}{2}}}$

$\sf\purple{\tt{Hence, \ the \ probability \ that \ the \ center}}$

$\sf\purple{\tt{in \ closed \ in \ figure \ is \ \dfrac{1}{2}.}}$

Answer 2

AnSwEr:

The probability is, in fact,

$\frac{1}{4}$

Wherever the first point is chosen, the diameter on which it lies (the diameter being determined by the circle center and the first chosen point) divides the circle into two symmetric semi-circles, so the second and third points (assuming they are distinct) must necessarily be place on opposite halves of the circle.

The line connecting the second and third points must then also lie above the center (with respect to the first point - or below the center, if the first point was on "top"); so if the second point is at a distance x from the first point along the perimeter of the circle (in units of the length of the perimeter), there's a range within length

$\frac{1}{2} - x$

in which to place the third point. Thus, computing the probability gives:

$2∫0 \: \frac{1}{2} (12−x)dx=2∫0 \: \frac{1}{2} x \: \: \: dx = \frac{1}{4} </p><p>$

where we multiply the integral by 2 to cover the fact that we can interchange the second and third point.