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?

**Answer**

The probability is \frac{1}{2} because the last flip determines it.

**Attribution***Source : Link , Question Author : Stuy , Answer Author : Asinomás*