A poll is given, showing $20\mathrm{\%}$ 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: $$P(X=k)=C(n,k)\ast (p^k)\ast ((1-p{)}^{(}n-k))$$ 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: $$P(X=1)=C(6,1)\ast ({0.2}^1)\ast ((1-0.2{)}^{(}6-1))$$

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 $\boxed{{\displaystyle 0.393}}$.