Understanding the Problem:
We are given that a fair dice is rolled twice and we need to find the probability that an odd number will follow an even number.

Approach:
To solve this problem, we can use the concept of conditional probability. We know that the probability of an event A occurring given that event B has already occurred is given by P(A|B) = P(A ∩ B) / P(B), where P(A ∩ B) represents the probability of both events A and B occurring together, and P(B) represents the probability of event B occurring.

Calculating the Probability:
Let's consider event A as the occurrence of an odd number and event B as the occurrence of an even number in the first roll.

We know that the probability of rolling an odd number on a fair dice is 3/6 = 1/2 (since there are 3 odd numbers out of 6 possible outcomes). Similarly, the probability of rolling an even number on a fair dice is also 1/2 (since there are 3 even numbers out of 6 possible outcomes).

Now, let's calculate the probability of both events A and B occurring together (P(A ∩ B)). Since the outcomes of the two rolls are independent, the probability of both events occurring together is equal to the product of their individual probabilities. Therefore, P(A ∩ B) = P(A) * P(B) = (1/2) * (1/2) = 1/4.

Next, let's calculate the probability of event B occurring (P(B)), which is the probability of rolling an even number on the first roll. As mentioned earlier, this probability is 1/2.

Finally, we can use the formula for conditional probability to find the probability of event A occurring given that event B has already occurred. Using the formula P(A|B) = P(A ∩ B) / P(B), we have P(A|B) = (1/4) / (1/2) = 1/2.

Therefore, the probability that an odd number will follow an even number is 1/2, which corresponds to option 'D' in the given choices.