Let $A$ be the event in which the selected student has opted for $NCC$ and $B$ be the event in which the selected student has opted for $NSS$.

Total number of students $=60$

Number of students who have opted for $NCC =30$

$\therefore $ $P(A)=\frac{30}{60}=\frac{1}{2}$

Number of students who have opted for $NSS =32$

$\therefore $ $P(B)=\frac{32}{60}=\frac{8}{15}$

Number of students who have opted for both $NCC$ and $NSS = 24$

$\therefore $ $P ( A$ and $B )=\frac{24}{60}=\frac{2}{5}$

The given information can be represented by a Venn diagram as

It is clear that Number of students who have opted for $NSS$ but not $NCC$

$=n(B-A)=n(B)-n(A \cap B)=32-24=8$

Thus, the probability that the selected student has opted for $NSS$ but not for $NCC$ 

$=\frac{8}{60}=\frac{2}{15}$