Let the event that a tree has disease be D and the event that tree is being damaged by insects be I.
Therefore,
[tex]\begin{gathered} |D|=54 \\ |I|=30 \\ |D\cup I|=68 \\ \end{gathered}[/tex]Recall that:
[tex]\begin{gathered} |D\cup I|=|D|+|I|-|D\cap I| \\ 68=54+30-|D\cap I| \\ \text{ Therefore,} \\ |D\cap I|=84-68=16 \end{gathered}[/tex]The required probability is given by:
[tex]\frac{|D\cap I|}{80}=\frac{16}{80}=20\%[/tex]Hence, the required probability is 20%=0.2
