How to Solve Probability Foundations — General probability — product-rule counting Questions on Exam P
Sample Practice Problem
ID: #Z67CV
Each of 4 independently sampled components is type \(A\), \(B\), or \(C\) with probabilities 0.5, 0.3, and 0.2. Let \(N_{A},N_{B},N_{C}\) be the respective type counts. Calculate the probability that the number of type \(C\) components exceeds the number of type \(A\) components by at least 2.
📖 Worked Solution & Strategy
The counts \((N_{A}, N_{B}, N_{C})\) follow a \(\operatorname{Multinomial}(4; 0.5, 0.3, 0.2)\) distribution with mass function \(\Pr(N_{A}=i, N_{B}=j, N_{C}=k) = \frac{4!}{i!j!k!}(0.5)^i(0.3)^j(0.2)^k\) for \(i+j+k=4\). The condition \(N_{C} \ge N_{A} + 2\) is satisfied by the valid count triples \((N_{A},N_{B},N_{C}) \in \{ (0,0,4), (0,1,3), (0,2,2), (1,0,3) \}\). Summing the multinomial probabilities over these specific triples gives \(0.0488\).