Final answer:
To ensure more than an 82% chance of a successful outcome, the inequality to be satisfied for the minimum number of trials n in a binomial experiment is ((9/11)^n) < 0.18. By setting up the inequality and solving for n using logarithms, we find the smallest integer that meets the requirement.
Step-by-step explanation:
The question asks for the minimum number of times that a binomial experiment must be run to ensure more than an 82% chance of success. Since the experiment is successful 2 out of every 11 times, we have a probability of success, p = 2/11, and a probability of failure, q = 1 - p = 9/11. A binomial experiment is defined by a fixed number of trials with two outcomes: success (p) and failure (q), with each trial being independent and conducted under identical conditions.
To solve this, we need to find the smallest number of trials, n, such that the probability of one or fewer failures (which results in at least one success) is less than 18% (because we want more than 82% chance of success).
Let's denote the random variable X as the number of successes in n trials. The inequality to satisfy is P(X ≥ 1) > 0.82. We can calculate the probability of having zero successes and subtract that from 1 to find the probability of having at least one success:
1 - P(X = 0) > 0.82,
which simplifies to:
1 - ((9/11)^n) > 0.82,
hence:
((9/11)^n) < 0.18
From here, we would use logarithms to solve for n. This would give us the smallest positive integer that satisfies the inequality, ensuring more than an 82% chance of at least one success in the experiment.