If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?

So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.

We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$

$n = 9, k = 0$

$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$ 

$n = 9, k = 2$

$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$ 

$n = 9, k = 4$
$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$

$n = 9, k = 6$

$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$

$n = 9, k = 8$

$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$

Add all of these up: 

$$=.64$$ so there's a 64% chance of probability?
The probability is $\frac{1}{2}$ because the last flip determines it.
Here's an analytical answer with greater emphasis on reasoning than anything specific to probability which might lend greater insight into the problem.

Consider if there was only one coin. The probability to have an even number of heads is $1\over2$, since there are two possible outcomes and we are only interested in one of them.

Now, let there be $N \gt 1$ coin tosses. The $N^{\text{th}}$ coin toss will either give a heads or tails. If the $N-1$ tosses resulted in an even number of heads the probability of the $N$ coin tosses resulting in an even number of heads is $1\over2$ since the $N^{\text{th}}$ coin toss will add either $0$ or $1$ to the number of heads from the $N-1$ tosses, and we are only concerned with the parity of the count. 

The conclusion is the same for the $N-1$ coin tosses resulting in an odd number of heads.

This approach is valid as this reasoning applies to all possible values of $N$ in our given domain.

$\therefore$ The probability of $N$ coin tosses resulting in an even number of heads is $1\over2$, with $N \in \mathbb{N}$.