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Let ${x(t),t>=0}$ be a compound poisson process with $x(t)=sum$ of $\xi$ between $1$ and $N(t)$, and suppose that $\lambda=1$ and $P{\xi=j}=j/10$ ,$j=1,2,3,4$. Calculate $p{x(n)=20}$.

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Here is a possible start, using the better notation $X(t) = \sum\nolimits_{i = 1}^{N(t)} {X_i }$. By the law of total probability, conditioning on $N(n)$,

$$ {\rm P}(X(n) = 20) = \sum\limits_{k = 0}^\infty {{\rm P}(X(n) = 20|N(n) = k){\rm P}(N(n) = k)}. $$ Hence, since the $X_i$ take values in $\{1,2,3,4\}$, $$ {\rm P}(X(n) = 20) = \sum\limits_{k = 5}^{20} {{\rm P}(X_1 + \cdots + X_k = 20)\frac{{e^{ - n} n^k }}{{k!}}}. $$

Another possibility. Use the probability-generating function of a compound Poisson process (from which you can obtain the probability mass function of $X(n)$ upon differentiation).

EDIT: The probability-generating function of $X(n)$ is given by $$ G_n (z):={\rm E}[z^{X(n)} ] = e^{n[P_{X_i } (z) - 1]} , $$ where $P_{X_i} (z) = \sum\nolimits_{j = 1}^4 {z^j \frac{j}{{10}}} $ is the probability-generating function of $X_i$ (standard exercise). Hence, using that $$ {\rm P}(X(n) = k) = \frac{{G_n^{(k)} (0)}}{{k!}} $$ (where $G_n^{(k)} (0)$ is the $k$th derivative evaluated at $0$), you can in principle obtain the probability in question, that is ${\rm P}(X(n) = 20)$.

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