Here, U = {1, 2, 3, …, 120} ‘

A number is selected at random.

∴ Total number of primary outcomes is n = ¹²⁰C₁ = 120.

(1) A = Event that the number selected is a multiple of 3.

= {3, 6, 9, 12, …, 117, 120}

∴ Favourable outcomes for the event A is m = 40.

Hence, P(A) = \(\frac{m}{n} = \frac{40}{120} = \frac{1}{3}\)

(2) A’ = Event that the number selected is not a multiple of 3.

∴ P(A’) = 1 – P(A)

= \(1 – \frac{1}{3} = \frac{2}{3}\)

(3) B = Event that the number selected is a multiple of 4.

= {4, 8, 12, 16, …, 116, 120}

∴ Favourable outcomes for the event B is m = 30.

Hence, P(B) = \(\frac{m}{n} = \frac{30}{120} = \frac{1}{4}\)

(4) B’ = Event that the selected number is not a multiple of 4.

∴ P(B’) = 1 – P(B)

= 1 – \(\frac{1}{4} = \frac{3}{4}\)

(5) A ∩ B = Event that the number selected is a multiple of both 3 and 4, i.e., a multiple of 12.

∴ A ∩ B = {12, 24, 36, 48, …, 108, 120}

∴ Favourable outcomes for the event A ∩ B is m = 10.

Hence, P(A ∩ B) = \(\frac{m}{n} = \frac{10}{20} = \frac{1}{12}\)