Two integers between 1 and 100, inclusive, each randomly and independe

Total number of integers = 100
Number of Even integers = 50
Number of Odd integers = 50

Probability of selecting an odd integer = \(\frac{50}{100}\) = \(\frac{1}{2}\)
Probability of selecting an even integer = \(\frac{50}{100}\) = \(\frac{1}{2}\)

From number properties, we know that for addition of two integers to be even => Odd + Odd = Even or Even + Even = Even

Using the above cases, the probability that the addition of two integers is even (Pa)= \([\frac{1}{2} * \frac{1}{2}] + [\frac{1}{2} * \frac{1}{2}]\) = \(\frac{1}{4} + \frac{1}{4}\) = \(\frac{1}{2}\)

Similarly , from number properties, we know that for multiplication of two integers to be even => Even * Odd = Even or Odd * Even = Even or Even * Even = Even

Using the above cases, the probability that the multiplication of two integers is even (Pm)= \([\frac{1}{2} * \frac{1}{2}] + [\frac{1}{2} * \frac{1}{2}] + [\frac{1}{2} * \frac{1}{2}]\) = \(\frac{1}{4} + \frac{1}{4} + \frac{1}{4}\) = \(\frac{3}{4}\)

From the question stem , we know that both addition and multiplication have an equal chance i.e. \(\frac{1}{2}\) * Pa and \(\frac{1}{2}\) * Pm

Therefore, the total Probability for the result to be even = \(\frac{1}{2}\) * Pa + \(\frac{1}{2}\) * Pm = \([\frac{1}{2} * \frac{1}{2}] + [\frac{1}{2} * \frac{3}{4}]\) = \(\frac{1}{4} + \frac{3}{8}\) = \(\frac{5}{8}\)

Answer is (C)