Question
Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with three vertices is equilateral, equals
A.
$$\frac{1}{2}$$
B.
$$\frac{1}{5}$$
C.
$$\frac{1}{10}$$
D.
$$\frac{1}{20}$$
Answer :
$$\frac{1}{10}$$
Solution :
Out of 6 vertices 3 can be chosen in $$^6{C_3}$$ ways.
$$\Delta $$ will be equilateral if it is $$\Delta \,ACE$$ or $$\Delta \,BDF$$ (2 ways)
∴ Required prob. $$ = \frac{2}{{^6{C_3}}} = \frac{2}{{20}} = \frac{1}{{10}}$$
Out of 6 vertices 3 can be chosen in $$^6{C_3}$$ ways.
$$\Delta $$ will be equilateral if it is $$\Delta \,ACE$$ or $$\Delta \,BDF$$ (2 ways)
∴ Required prob. $$ = \frac{2}{{^6{C_3}}} = \frac{2}{{20}} = \frac{1}{{10}}$$