Answer: (1) 4/5 (2) 1/5
Tens digit: 5 choices, units digit: 4 remaining choices, so n(S) = \(5 \times 4 = 20\).
S = {23, 25, 27, 29, 32, 35, 37, 39, 52, 53, 57, 59, 72, 73, 75, 79, 92, 93, 95, 97}
(1) The number is odd unless its units digit is 2. Numbers ending in 2: 32, 52, 72, 92 (4 numbers). So n(A) = 20 − 4 = 16.
\[ P(A) = \frac{16}{20} = \frac{4}{5} \]
(2) Multiples of 5 end in 5: B = {25, 35, 75, 95}, n(B) = 4.
\[ P(B) = \frac{4}{20} = \frac{1}{5} \]