4. (1988 American High School Mathematics Examination) Toss 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 $p$ value such that $w=\frac{144}{625}$
4. D. Reason: Let the probability of the reverse side facing up be $q$, then $q=1-p$.

In 5 independent tosses, the probability of exactly 3 heads facing up should be 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 fifth-degree equation in $p$, let the left side be $f(p)$.
Since $f(0)=f(1)=-\frac{3^{2} \cdot 2^{3}}{5^{2}}0$,
$f(p)$ has real roots in $\left(0, \frac{1}{2}\right),\left(\frac{1}{2}, 1\right)$, so the value of $p$ is not unique.