<s>[INST] 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?
 [/INST] There are two cases here:


*

*There's an even number of heads: 0, 2, 4, 6 or 8 heads

*There's an odd number of heads: 1, 3, 5, 7 or 9 heads


But notice that having an odd number of heads means having an even number of tails (e.g. 5 heads means 4 tails), so the second case is the same as:


*

*There's an even number of tails: 0, 2, 4, 6 or 8 tails


Since heads and tails are equally probable, we can, by symmetry, see that these two cases have the same probability. Therefore each must have probability $1/2$.
</s>