Asked in: June 2026 board paperQ.6 A two digit number is formed using digits 2, 3, 5, 7, 9 without repetition. What is the probability that the number formed is:(a) an odd number.(b) a multiple of 5.

Answer: (a) \(\dfrac{4}{5}\) (b) \(\dfrac{1}{5}\)

S = {23, 25, 27, 29, 32, 35, 37, 39, 52, 53, 57, 59, 72, 73, 75, 79, 92, 93, 95, 97}, n(S) = 20

(Tens digit: 5 choices, units digit: 4 remaining choices, 5 × 4 = 20.)

(a) Event A: odd number, so units digit is 3, 5, 7 or 9. Every number except those ending in 2 is odd. Numbers ending in 2: 32, 52, 72, 92 (4 numbers). So n(A) = 20 − 4 = 16.

\(P(A) = \dfrac{16}{20} = \dfrac{4}{5}\)

(b) Event B: multiple of 5, so units digit is 5. B = {25, 35, 75, 95}, n(B) = 4

\(P(B) = \dfrac{4}{20} = \dfrac{1}{5}\)