I recently had an interview question that posed the following... Suppose you are shooting free throws and each shot has a 60% chance of going in (there are no "learning" or "depreciation" effects, all have the some probability no matter how many shots you take).
Now there are three scenarios where you can win $1000
- Make at least 2 out of 3
- Make at least 4 out of 6
- Make at least 20 out of 30
My initial thought is that each are equally appealing as they all require the same percentage of free throw shots. However when using a binomial calculator (which this process seems to be) the P (X > x) seems to be the highest for scenario 1. Is this due to the number of combinations?