Answer
$P(E)=\frac{N(E)}{N(S)}=\frac{3}{8}$
Work Step by Step
H -> head
T -> tail
All the possible outcomes (sample space):
$S=[(H,H,H),(H,H,T),(H,T,H),(H,T,T),(T,H,H),(T,H,T),(T,T,H),(T,T,T)]$
We have that the total number of possible outcomes in the sample space is:
$N(S)=8$
We want exactly two tails (Event):
$E=[(H,T,T),(T,H,T),(T,T,H)]$
$N(E)=3$
$P(E)=\frac{N(E)}{N(S)}=\frac{3}{8}$
