How to Solve Probability Foundations — General probability — counting / sampling without replacement Questions on Exam P
Sample Practice Problem
ID: #UGIAM
An urn contains 3 white, 3 red, and 2 black balls. Three draws are made with replacement. Calculate the probability all three draws have different colors.
📖 Worked Solution & Strategy
Total urn capacity is \(T = 3+3+2 = 8\). Independent single-draw color probabilities are \(p_{W} = 3/8\), \(p_{R} = 3/8\), and \(p_{B} = 2/8\). For three different colors, multiply product of single-draw probabilities by \(3! = 6\) possible orderings: \(\Pr(\text{all diff}) = 6 p_{W} p_{R} p_{B} = 6\left(\dfrac{3}{8}\right)\left(\dfrac{3}{8}\right)\left(\dfrac{2}{8}\right)\). Therefore, probability all three draws have different colors is \(\frac{27}{128}\). This is the correct option.