$14 \cdot 28$ Tossing an "unfair" coin, the probability of heads is $p$. Let $w$ be the probability of getting exactly 3 heads in 5 independent tosses. If $w=\frac{144}{625}$, then
(A) $p$ must be $\frac{2}{5}$.
(B) $p$ must be $\frac{3}{5}$.
(C) $p$ must be greater than $\frac{3}{5}$.
(D) $p$ is not uniquely determined.
(E) There is no value of $p$ that makes $w=\frac{144}{625}$.
(39th American High School Mathematics Examination, 1988)
［Solution］Let the probability of the coin landing tails up be $q$, then
$$
q=1-p \text {. }
$$

In 5 independent tosses, the probability of exactly 3 heads up is the term containing $p^{3}$ in the expansion of $(p+q)^{5}$.
The coefficients of its expansion are $1,5,10,10,5,1$, so the required term is
$$
10 p^{3} q^{2}=10 p^{3}(1-p)^{2} \text {. }
$$

That is, $\frac{144}{625}=10 p^{3}(1-p)^{2}$ or $p^{3}(1-p)^{2}-\frac{2^{3} \cdot 3^{2}}{5^{5}}=0$.
This is a quintic equation in $p$, let the left side be $f(p)$.
$\because f(0)=f(1)=-\frac{3^{2} \cdot 2^{3}}{5^{2}}0$,
$\therefore f(p)$ has real roots in $\left(0, \frac{1}{2}\right),\left(\frac{1}{2}, 1\right)$.
Thus, the value of $p$ is not unique.
Therefore, the correct choice is $(D)$.