Question

There are 10 points in a plane no three of which are in the same straight line, excepting 4 points, which are collinear.

Find the probability that

(i) number of straight lines obtained from the pairs of these points.

(ii) number of triangles that can be formed with the vertices as these points.

Answer 1

(i) Number of straight lines formed joining the 10 points taking 2 at a time.

$n\left(S\right)={}^{10}{C}_2=\frac{10!}{2!8!}=45$

Now, number of straight lines formed by joining the four points taking 2 at a time

$={}^4{C}_2=\frac{4!}{2!2!}=6$

But 4 collinear points, when joined pairwise gives only one line.

$\therefore$ n(E) = 45 - 6 + 1 = 40

$\therefore$ Required probability $=\frac{n\left(E\right)}{n\left(S\right)}=\frac{40}{45}=\frac{8}{9}$

(ii) Number of triangles formed by joining the points taking 3 at a time,

$n\left(S\right)={}^{10}{C}_3=\frac{10!}{3!7!}=120$

Number of triangles formed by joining the 4 points taken 3 at a time
$={}^4{C}_3=\frac{4!}{3!1!}=4$

But 4 collinear points cannot form a triangle, when taken 3 at a time. So, the required triangles

n(E) = 120 - 4 = 116

$\therefore$ Required probability $=\frac{n\left(E\right)}{n\left(S\right)}=\frac{116}{120}=\frac{29}{30}$