Correct answer is D.
The sum will be even when; both numbers are either even or odd,
i.e. for both numbers to be even, the total cases \(5_{C_1}\times 4_{C_1}\) (Both the numbers are odd) + \(4_{C_1}\times 3_{C_1}\) (Both the numbers are even)= 32
The favourable number of cases will be,
Both odd, i.e. selecting numbers from 1, 3, 5, 7, or 9, i.e.
\(5_{C_1}\times 4_{C_1}\) = 20
Thus, the probability that both numbers are odd will be
\(=\frac{Favorable\,outcomes}{Total\,outcomes}\)
\(\Rightarrow\)\(\frac{20}{32}=\frac{5}{8}\)
