Solving Random Functions Of Other Random Functions

Question

Knowing that:

$$X = B(N, P)$$ $$P = \frac{1}{2}$$ $$N = ODD$$

Essentially, knowing that X is a binomial distribution with parameters of [P = $\frac{1}{2}$, N = Any Odd Number], I also know that $$Y\space = \space X \mod 2$$

Knowing this information, I'm trying to go about calculating: $$P(Y = 1)$$

From an intuitive sense, I believe the answer should be $\frac{2}{3}$ simply because, if there are N trials, and N is an odd number, it's easy to assume that 2 out of the 3 numbers would be odd, satisfying that $X \mod 2 = 1$. From a mathetmical standpoint, I was thinking of doing:

$$P(Y = 1) = \sum_{n=1}^{\infty} {n \choose 1}\cdot \frac12\cdot(1-\frac12)$$

I understand my summation is wrong; however, I cannot think of what logic to use when coming up with this equation.

Answer 1

Imagine tossing a coin once. Can you see that the probability of getting zero heads is equal to the probability of getting one head? So what is the probability of getting an odd number of heads when you toss the coin once?

Now imagine tossing a coin three times. Can you see that the probability of getting zero heads is equal to the probability of getting three heads? And can you see that the probability of getting one head is equal to the probability of getting two heads? So what is the probability of getting an odd number of heads when you toss the coin three times?

Now imagine tossing a coin five times...