Question

There are 6 points in a plane out of which 3 points are collinear. What is the probability if 3 points which are selected form a triangle?

• A. (19/20)
• B. (20/21)
• C. (17/20)
• D. (19/22)

Answer 1

The correct option is A

(19/20)

Let A,B,C,D,E,F be the six points in which B,E, and F are collinear.
The numbe rof outcomes i.e., selection of 3 points =
(A,B,C), (A,B,D), (A,B,E), (A,B,F), (A,C,D), (A,C,E), (A,C,F), (A,D,E), (A,D,F), (A,E,F), (B,C,D), (B,C,E), (B,D,E), (B,D,F), (B,E,F), (C,D,E), (C,D,F), (C,E,F), (D,E,F) = 20
There is only possible outcome where a triangle cannot b eformed because all three points are collinea = (B,E,F)
Total number of favorable outcomes = 19
$Probability=\frac{Totalnumberoffavourableoutcomes}{Totalnumberofoutcomes}$
$=\frac{19}{20}$