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 \(0.0488\). The evaluated result is \(0.0488\), which matches the correct option.