Solution:
Given:
[tex]\begin{gathered} Sample\text{ space}=\lbrace15,16,17,18,19,20,21,22,23\rbrace \\ Total\text{ n}umber\text{ of possible outcomes}=9 \\ \\ \\ Numbers\text{ divisible by 3}=\lbrace15,18,21\rbrace \\ Number\text{ of required outcome}=3 \end{gathered}[/tex]Therefore, the probability of picking a number divisible by 3 is;
[tex]\begin{gathered} P(a\text{ number divisible by 3\rparen}=\frac{3}{9} \\ \\ Simplifying\text{ further,} \\ P(a\text{ number divisible by 3\rparen}=\frac{1}{3} \end{gathered}[/tex]