Question

$18$ points are indicated on the perimeter of a triangle $ABC$ (see figure). If three points are choosen probability that it will form a triangle.

• A. $\frac{331}{816}$
• B. $\frac{1}{2}$
• C. $\frac{355}{408}$
• D. None of these

Answer 1

The correct option is D None of these

Given that,

18 points are indicated on the parameter of a $\Delta ABC$.

Then, five points on a line. But total of 7 points on that line.

According to given question

Total number of triangle ${=}^{18}{C}_3$

But triangle can not be formed by joining the point which are in the straight line.

So possible triangle

$=\frac{{}^{18}{C}_3-\left({}^7{C}_3{+}^7{C}_3{+}^7{C}_3\right)}{{}^{18}{C}_3}$

$=\frac{{}^{18}{C}_3-3{\times }^7{C}_3}{{}^{18}{C}_3}$

$=\frac{\frac{18!}{3!(18-3)!}-3\times \frac{7!}{3!(7-3)!}}{{}^{18}{C}_3}$

$=\frac{\frac{18!}{3!15!}-3\times \frac{7!}{3!4!}}{{}^{18}{C}_3}$

$=\frac{\frac{18\times 17\times 16\times 15!}{3\times 2\times 1\times 15!}-3\times \frac{7\times 6\times 5\times 4!}{3\times 2\times 1\times 4!}}{{}^{18}{C}_3}$

$=\frac{48\times 17-3\times 35}{816}$

$=\frac{711}{816}$

Hence, this is the answer.