Answer
No, $P(more~likely~|~18~to~34)\lt P(more~likely)$.
Work Step by Step
N(18 to 34) = 542 and N(18 to 34 and more lilkely) = 238.
$P(more~likely~|~18~to~34)=\frac{N(18~to~34~and~more~lilkely)}{N(18~to~34)}=\frac{238}{542}\approx0.439$
The sample space: 2160 randomly selected adult americans. So, N(S) = 2160. And N(more likely) = 1329. So:
$P(more~likely)=\frac{N(more~likely)}{N(S)}=\frac{1329}{2160}=0.615$
We conclude that:
$P(more~likely~|~18~to~34)\lt P(more~likely)$.
