Condition on the fair coin choosing the initial chamber. The law of total probability gives $\mathbb{P}(\text{reach }k)=(p_1+p_2)/2$. With $r=(1-p)/p$, substituting the biased gambler's ruin probability givesBoth numerator and denominator in the first case have the same sign. Its limit as $r\to 1$ is the fair-walk value, since the ratio of their first derivatives is $3/(2k)$. The averaging is over initial locations, not over the transition probabilities.

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 $0.9^{10}$. HenceThe 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.