$14 \cdot 29$ A biased coin has a probability of $\frac{2}{3}$ of landing heads when flipped. If the coin is flipped 50 times, then the probability that the number of heads is even is
(A) $25\left(\frac{2}{3}\right)^{50}$.
(B) $\frac{1}{2}\left(1-\frac{1}{3^{50}}\right)$.
(C) $\frac{1}{2}$.
(D) $\frac{1}{2}\left(1+\frac{1}{3^{50}}\right)$.
(E) $\frac{2}{3}$.
(43rd American High School Mathematics Examination, 1992)
[Solution] Clearly, for 50 coin tosses, the probability of the number of heads being even is $\left(\frac{1}{3}+\right.$ $\left.\frac{2}{3}\right)^{50}$, which is the sum of the terms in the expansion where the power of $\frac{2}{3}$ is even, denoted as $p$.
By $1=\left(\frac{1}{3}+\frac{2}{3}\right)^{50}=\frac{1}{3^{50}}(1+2)^{50}$,
and
$$
(1-2)^{50}=(-1)^{50}=1 \text {, }
$$

the sum of the even powers of -2 and the sum of the odd powers of -2 (in absolute value) in its expansion is 1. Thus, in the $\circledast$ expression, the sum of the even powers of 2, $S_{1}$, and the sum of the odd powers of 2, $S_{2}$, in the expansion of $(1+2)^{50}$ also differ by 1.
Therefore, $p=\frac{1}{2}\left\{\frac{1}{3^{50}}\left[\left(S_{1}+S_{2}\right)+1\right]\right\}=\frac{1}{2}\left(1+\frac{1}{3^{50}}\right)$. Hence, the answer is $(D)$.