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) = \(\frac{m}{n} = \frac{10}{20} = \frac{1}{12}\)