The probability of no heads is given by:
\(P(0\ heads)=3C0\times0.5^{0}\times0.5^{3}=\frac{1}{8}\)
The probability of exactly one head is given by:
\(P(1\ head)=3C1\times0.5^{1}\times0.5^{2}=\frac{3}{8}\)
Therefore it is three times as likely that one head will appear as that no head will appear.
The statement in the question is FALSE.
