# 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?

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

$$n = 9, k = 0n = 9, k = 0$$

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

$$n = 9, k = 2n = 9, k = 2$$

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

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

$$n = 9, k = 6n = 9, k = 6$$

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

$$n = 9, k = 8n = 9, k = 8$$

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

Add all of these up:

$$=.64=.64$$ so there’s a 64% chance of probability?

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