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?

Solution 1:

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

Solution 2:

If there are an even number of heads then there must be an odd number of tails. But heads and tails are symmetrical, so the probability must be $1/2$.

Solution 3:

Your approach is good also, you probably made a mistake in calculations. The number of favorable outcomes is $$\binom{9}{0}+\binom{9}{2}+\binom{9}{4}+\binom{9}{6}+\binom{9}{8}=1+36+126+84+9=256$$ The number of all possible outcomes is $512$ thus the probability to get even number of heads is $0.5$.

Solution 4:

The easiest way to see this : Consider the number of heads we have in the first $8$ coins.

• If this number is even, we need a tail, we have probability $\frac{1}{2}$
• If this number is odd, we need a head, we have probability $\frac{1}{2}$

Hence no matter what the $8$ coins delivered, we have probability $\frac{1}{2}$ , which is the answer.

Solution 5:

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 $1/2$.