An ordinary dice is rolled for a certain number of times. If the probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times, then the probability of getting an odd number for odd number of times is:

We are told that a dice is rolled $$n$$ times. The probability of getting an odd number on a single roll is $$\frac{1}{2}$$ and similarly for an even number.

The probability of getting an odd number exactly $$k$$ times follows a binomial distribution: $$P(X = k) = \binom{n}{k} \left(\frac{1}{2}\right)^n$$.

We are given that the probability of getting an odd number 2 times equals the probability of getting an even number 3 times. So $$\binom{n}{2}\left(\frac{1}{2}\right)^n = \binom{n}{3}\left(\frac{1}{2}\right)^n$$.

This simplifies to $$\binom{n}{2} = \binom{n}{3}$$, that is, $$\frac{n(n-1)}{2} = \frac{n(n-1)(n-2)}{6}$$.

Dividing both sides by $$\frac{n(n-1)}{2}$$, we get $$1 = \frac{n - 2}{3}$$, so $$n = 5$$.

Now we need the probability of getting an odd number an odd number of times, i.e., $$P(X = 1) + P(X = 3) + P(X = 5)$$ with $$n = 5$$.

$$= \left(\frac{1}{2}\right)^5 \left[\binom{5}{1} + \binom{5}{3} + \binom{5}{5}\right] = \frac{1}{32}(5 + 10 + 1) = \frac{16}{32} = \frac{1}{2}$$

Hence, the correct answer is Option D.