μ = 30 n(W) = 15 n(M) = 13 n(W∪M)1=6 Let x = number of students that study both woodwork and metalwork i.e. n(W ∩ M) = x Number of students that study only woodwork,n(W∩M1) = 15−x Number of students that study only metalwork, n(W1∩M) = 13−x Bringing all together, n(W∩M1) +n(W1∩M) + n(W∩M) + n(W∪M)1 = μ ∴ (15 - x) + (13 - x) + x + 6 = 30 ⇒ 34 - x = 30 ⇒ 34 - 30 = x ∴ x = 4 n(W∩M1) = 15−4=11 ∴ The number of students that study woodwork but not metalwork is 11.