A compound of formula $A_2{B}_3$ has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms.

Here  can also be written as  $A_4{B}_6$

Thus , hcp has six atoms so, ‘B’ forms hcp lattice and ‘A’ is present in tetrahedral void.

Total tetrahedral voids =12

Fraction of tetrahedral voids occupied by $A=\frac{4}{12}=\frac{1}{3}$