= Solution
Let $A_i$ be the <event> that candidate $i$ succeeds. The <countable additivity> axiom adds <probabilities> only for mutually disjoint <events>. Successes of different candidates are not mutually exclusive, so replacing the <probability> of their <union> by $\sum_i\mathbb P(A_i)=1$ misuses that axiom. Without disjointness one obtains only the <union bound>, not certainty. The same sum is the <expected value> of the number of successes, which also need not imply at least one.
If the ten outcomes are mutually <independent events>, with each success <probability> $0.1$, all fail with <probability> $0.9^{10}$. Hence
$$
\boxed{\mathbb P(\text{at least one success})=1-0.9^{10}
=0.6513215599\ldots.}
$$
The <independence> assumption is essential. The marginal <probabilities> alone allow any <union> <probability> from $0.1$ up to $1$; perfectly overlapping success <events> give the lower endpoint and disjoint successes give the upper endpoint.
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