Final answer:
The probability that all five citizen-entities from a sample are of pension age, given that the individual probability is 0.6, is about 0.07776 when rounded to four decimal places.
Step-by-step explanation:
About 0.6 is the probability that a randomly selected Cygnus citizen-entity is of pension age. To find the probability that in a randomly selected sample of five citizen-entities, all of them are of pension age, we can use the binomial probability formula since each selection is independent, and there are only two outcomes (of pension age or not).
The probability of all five being of pension age is given by P(X = 5) where X is the number of citizen-entities of pension age. The formula for this is:
P(X = k) = nCk * p^k * (1-p)^(n-k) where n is the number of trials, k is the number of successful outcomes we want, p is the probability of success on a single trial, and nCk represents the binomial coefficient.
In this case, n = 5, k = 5, and p = 0.6. The binomial coefficient nCk is 1 when k = n, so it simplifies to:
P(X = 5) = 1 * 0.6^5 * (1-0.6)^0 which calculates to P(X = 5) ≈ 0.07776.
Therefore, the probability that all five citizen-entities are of pension age is about 0.07776, rounded to four decimal places.