smc

We have $w=\left(\genfrac{}{}{0ex}{}{5}{3}\right)p^3(1-p{)}^2=10p^3(1-p{)}^2$, so we need to solve $10p^3(1-p{)}^2=\frac{144}{625}⟹p^3(1-p{)}^2=\frac{72}{3125}$. Now observe that when $p=0$, the left-hand side evaluates to $0$, which is less than $\frac{72}{3125}$; when $p=\frac{1}{2}$, it evaluates to $\frac{1}{32}$, which is more than $\frac{72}{3125}$, and when $p=1$, it evaluates to $0$ again. Thus, since $p^3(1-p{)}^2$ is continuous, the Intermediate Value Theorem tells us there is a solution between $0$ and $\frac{1}{2}$ and another solution between $\frac{1}{2}$ and $1$, meaning that there is not a unique value of $p$, so the answer is $\boxed{p\text{ is not uniquely determined}}$.