Defective Items

Solution

Answer: 0.13

The machines operate independently. The total number of defective items is \[ T = X + Y. \] We want \[ P(T = 3). \] This can happen in several ways: \[ (X=0,Y=3), \quad (X=1,Y=2), \quad (X=2,Y=1), \quad (X=3,Y=0). \] So \[ P(T=3) = P(X=0,Y=3) + P(X=1,Y=2) + P(X=2,Y=1) + P(X=3,Y=0). \] Because \(X\) and \(Y\) are independent, \[ P(X=a,Y=b) = P(X=a)P(Y=b). \] Using the binomial formula \[ P(X=k) = \binom{5}{k}(0.1)^k(0.9)^{5-k} \] \[ P(Y=k) = \binom{5}{k}(0.2)^k(0.8)^{5-k}. \] Compute the required probabilities. \[ P(X=0) = (0.9)^5 = 0.59049 \] \[ P(X=1) = \binom{5}{1}(0.1)(0.9)^4 = 0.32805 \] \[ P(X=2) = \binom{5}{2}(0.1)^2(0.9)^3 = 0.0729 \] \[ P(X=3) = \binom{5}{3}(0.1)^3(0.9)^2 = 0.0081 \] For machine B \[ P(Y=0) = (0.8)^5 = 0.32768 \] \[ P(Y=1) = \binom{5}{1}(0.2)(0.8)^4 = 0.4096 \] \[ P(Y=2) = \binom{5}{2}(0.2)^2(0.8)^3 = 0.2048 \] \[ P(Y=3) = \binom{5}{3}(0.2)^3(0.8)^2 = 0.0512 \] Now multiply the combinations. \[ P(X=0,Y=3) = 0.59049 \times 0.0512 = 0.030233 \] \[ P(X=1,Y=2) = 0.32805 \times 0.2048 = 0.067184 \] \[ P(X=2,Y=1) = 0.0729 \times 0.4096 = 0.029860 \] \[ P(X=3,Y=0) = 0.0081 \times 0.32768 = 0.002654 \] Adding these probabilities, \[ P(T=3) = 0.030233 + 0.067184 + 0.029860 + 0.002654. \] \[ P(T=3) \approx 0.1299. \]