Answer: P(A) = 7/10, P(B) = 3/5
Let the boys be B1, B2, B3 and the girls G1, G2.
S = {B1B2, B1B3, B2B3, B1G1, B1G2, B2G1, B2G2, B3G1, B3G2, G1G2}, n(S) = 10.
A = all committees except the three all-boy ones, n(A) = 7, so \(P(A) = \dfrac{7}{10}\).
B = {B1G1, B1G2, B2G1, B2G2, B3G1, B3G2}, n(B) = 6, so \(P(B) = \dfrac{6}{10} = \dfrac{3}{5}\).