We can solve this problem by first finding the probability that none of the 5 people selected have been vaccinated, and then subtracting this from 1 to get the probability that at least one person has been vaccinated.

Steps:

1. Probability that a person is not vaccinated:
   If 69% of people are vaccinated, the probability that a person is not vaccinated is: $P(\text{not vaccinated}) = 1 - 0.69 = 0.31$

2. Probability that none of the 5 people are vaccinated:
   Since the selections are independent, the probability that all 5 people are not vaccinated is: $P(\text{none vaccinated}) = 0.31^5$

3. Probability that at least one person is vaccinated:
   The probability that at least one person is vaccinated is the complement of the probability that none are vaccinated: $P(\text{at least one vaccinated}) = 1 - P(\text{none vaccinated})$

Now, let's calculate this:

$P(\text{none vaccinated}) = 0.31^5$ $P(\text{at least one vaccinated}) = 1 - 0.31^5$

I'll calculate the exact value.The probability that at least one of the 5 randomly selected people has been vaccinated is approximately 0.9971 (to four decimal places).

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:

1. How would the probability change if 3 people were randomly selected instead of 5?
2. What if the vaccination rate were 80% instead of 69%?
3. How can this problem be generalized for $n$ people and a vaccination rate $p$?
4. What is the probability that exactly one person out of the 5 is vaccinated?
5. How does the assumption of independence affect this calculation?

Tip: Always use complements when calculating "at least one" probabilities, as it's often simpler.