A poll is given, showing $20 \%$ are in favor of a new building project.
If 6 people are chosen at random, what is the probability that exactly 1 of them favor the new building project?

Step 1 ：This problem is a binomial probability problem. The binomial distribution model deals with finding the probability of success of an event which has only two possible outcomes in a series of experiments. In this case, the two outcomes are whether a person is in favor of the new building project or not. We are looking for the probability that exactly 1 out of 6 people chosen at random favor the new building project.

Step 2 ：The formula for binomial probability is: \[P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))\] where: \[P(X=k)\] is the probability we are trying to find, \[C(n, k)\] is the number of combinations of n items taken k at a time, \[p\] is the probability of success (in this case, the probability that a person is in favor of the new building project), \[n\] is the number of trials (in this case, the number of people chosen at random), and \[k\] is the number of successes we want (in this case, the number of people who favor the new building project).

Step 3 ：In this case, n=6, k=1, and p=0.2. We can substitute these values into the formula to find the answer.

Step 4 ：Substituting the values into the formula, we get: \[P(X=1) = C(6, 1) * (0.2^1) * ((1-0.2)^(6-1))\]

Step 5 ：Solving the above expression, we get the probability as approximately 0.3932160000000002.

Step 6 ：Final Answer: The probability that exactly 1 out of 6 people chosen at random favor the new building project is approximately \(\boxed{0.393}\).