= Solution
Direct integration of the <uniform distribution> yields $\mathbb E U_j=b/2$ and $\mathbb E U_j^2=b^2/3$. The <expectation of a product of independent random variables> gives
$$
\mathbb EW_n=\prod_{j=1}^n\mathbb E U_j=(b/2)^n,\qquad \mathbb EW_n^2=(b^2/3)^n.
$$
Subtracting the square of the <expected value> gives the <variance>:
$$
\boxed{\mathbb EW_n=(b/2)^n,\qquad\operatorname{Var}(W_n)=(b^2/3)^n-(b^2/4)^n.}
$$
Since $W_n\geq0$, <Markov's inequality> implies, for every $\varepsilon>0$,
$$
\mathbb P(W_n>\varepsilon)\leq\frac{(b/2)^n}{\varepsilon}\longrightarrow0\quad(0<b<2).
$$
Hence $\boxed{W_n\to0\text{ in probability for }0<b<2.}$ The first <moment> suffices: the <variance> need not tend to zero throughout this range.
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