12. (1992 American High School Mathematics Examination) A biased coin lands heads up with probability $\frac{2}{3}$. If the coin is tossed 50 times, 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. $\frac{2}{3}$
12. D. Reason: Clearly, the probability of getting an even number of heads in 50 coin tosses is the sum of the terms with even powers of $\frac{2}{3}$ in the expansion of $\left(\frac{1}{3}+\frac{2}{3}\right)^{50}$, denoted as $p$.
By $1=\left(\frac{1}{3}+\frac{2}{3}\right)^{50}=\frac{1}{3^{50}}(1+2)^{50}$,
and $\quad(1-2)^{50}=(-1)^{50}=1$,
the sum of the absolute values of the terms with even powers of -2 and the terms with odd powers of -2 in its expansion is 1.
Thus, in (*) the sum of the terms with even powers of 2, $S_{1}$, and the sum of the terms with odd powers of 2, $S_{2}$, in the expansion of $(1+2)^{50}$ also differ by 1.
Therefore, $p=\frac{1}{2}\left\{\frac{1}{3^{50}}\left[\left(S_{1}+S_{2}\right)+1\right]\right\}=\frac{1}{2}\left(1+\frac{1}{3^{50}}\right)$.