Assume that a procedure yields a binomial distribution with a trial repeated $n=5$ times. Use some form of technology to find the probability distribution given the probability $p=0.15$ of success on a single trial.
(Report answers accurate to 4 decimal places.)

Step 1 ：Let's assume that a procedure yields a binomial distribution with a trial repeated 5 times. We are asked to find the probability distribution given the probability 0.15 of success on a single trial.

Step 2 ：The binomial distribution is a probability distribution that describes the number of successes in a sequence of n independent experiments. In this case, n=5 and p=0.15.

Step 3 ：The probability mass function of a binomial distribution is given by: \[P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\] where: \[P(X=k)\] is the probability of k successes in n trials, \[\binom{n}{k}\] is the number of combinations of n items taken k at a time, p is the probability of success on a single trial, and n is the number of trials.

Step 4 ：We can use this formula to calculate the probability distribution for k=0 to 5.

Step 5 ：By substituting n=5 and p=0.15 into the formula, we get the probability distribution as [0.4437, 0.3915, 0.1382, 0.0244, 0.0022, 0.0001].

Step 6 ：Final Answer: The probability distribution for a binomial distribution with n=5 trials and a probability of success p=0.15 is \(\boxed{[0.4437, 0.3915, 0.1382, 0.0244, 0.0022, 0.0001]}\).