Since the bag has 20 tickets, total number of cases n = 20
Let A: getting a multiple of 2
B: getting a multiple of 5
Event A has 10 favourable outcomes
A = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}
∴ P(A) = \(\frac{10}{20} =\frac{1}{2}\)
Event B has four favourable outcomes
B = {5, 10, 15, 20)
∴ P(B) = \(\frac{4}{20}\)
A ∩ B = {10,20]
∴ P(A ∩ B) = \(\frac{2}{20}\)
∴ P(multiple of 2 or 5 ) = P(A ∪ B) =P(A) + P(B) – P(A ∩ B)
= \(\frac{10}{20} + \frac{4}{20} - \frac{2}{20} = \frac{12}{20} = \frac{3}{5}\)
