Biased Coin

If the number of heads is even, then there will be 0, 2, or 4 heads among the five. Let $X$ be the number of heads among the 5 tosses.

If there are 0 heads, then all five tosses will be tails.

$P(X=0)=\left(\frac{1}{3}\right)^5=\frac{1}{243}$

If there are 2 heads, then the remaining 3 tosses will be tails. There are $\binom{5}{2}$ combinations of 2 heads and 3 tails.

$P(X=2)=\binom{5}{2}\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)^3=\frac{40}{243}$

If there are 4 heads, then the remaining toss will be tails. There are $\binom{5}{4}$ combinations of 4 heads and 1 tails.

$P(X=4)=\binom{5}{4}\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)=\frac{80}{243}$

Each of these events is mutually exclusive. Therefore, the probability that there is an even number of heads is:

$P(X=0)+P(X=2)+P(X=4)=\boxed{\frac{121}{243}}.$