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?

Answer

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