= Solution
By <Poisson thinning>, the first arrival times of the coupon types are independent <exponential distribution> $T_j$ of rates $\lambda p_j$. Completion requires all those arrivals, so $T=\max_jT_j$ and
$$
\boxed{\mathbb P(T<t)=\prod_{j=1}^m(1-e^{-\lambda p_jt})}.
$$
To relate time to count rigorously, let $E_k$ be the successive independent exponential waiting times of mean $1/\lambda$. The completion count $L$ depends only on the coupon labels, which are independent of all $E_k$. Therefore
$$
T=\sum_{k\ge1}E_k\mathbf1_{\{L\ge k\}},\qquad
\mathbb ET=\sum_{k\ge1}\frac1\lambda\mathbb P(L\ge k)=\frac{\mathbb EL}{\lambda}.
$$
Nonnegative integration justifies the sum, avoiding any unproved optional-stopping assertion. Thus the <Poissonized coupon completion time and count> identity is \b[$\lambda\mathbb ET=\mathbb EL$]. Changing variables $s=\lambda t$ in the tail integral gives
$$
\mathbb EL=\int_0^\infty\left[1-\prod_j(1-e^{-p_js})\right]ds,
$$
which depends only on the type probabilities, not $\lambda$. These expectations are finite when all $p_j>0$; a missing type makes both infinite and completion impossible.
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