Problem: You’re hosting a 30-person Thanksgiving and need someone to bring a can of cranberry sauce. For each person you ask, the probability that they end up bringing a Can o’ Cran is 0.1. If you ask all 30 people, what’s the probability you’ll get exactly one can?
Bonus: You need cranberry sauce, but you don’t want to get bogged down with a lot of it. How many people should you ask to maximize the chance that you get between 1 and 3 cans?
A Solution: This is an independent trials problem.. The probability of success is 1/10, the probability of failure is 9/10. So the probability of exactly 1 success in thirty trials is $30(\frac{1}{10})(\frac{9}{10})^{29} \approx 0.1413$.
For the bonus: The probability of getting between 1 and 3 cans of cranberries is maximized at approximately 0.7517 if you ask 18 people to bring a can. (Run the code below to see the computation.)