If X~G(0,01): What is the probability that: The first success is the fortieth trial? The first twelve trials will fail? ## Exercise > An automatic machine manufactures microcircuits one at a time and independently. Each microcircuit has a probability of $0, 01$ being defective. > > a) Calculate the probability that the first defective microcircuit in a working day is the fortieth that is made. > > b) Probability that the first twelve microcircuits manufactured are all correct. ## Solution Let X: "Number of circuits diagnosed as correct, before diagnosing the first circuit as defective", then $X$~$G(0.01)$, so $f_X(x)=0.99^x*0.01$ * a) $f_X(39) = 0.00675729...$ * b) $f_X(12) = 0.008863848...$ **Is my solution correct?**

a) Okay. $f_X(39)$ is the probability that the first defective circuit is the fortieth circuit.

b) Rethink this one. $f_X(12)$ is the probability that the first defective circuit is exactly the thirteenth circuit. You want $1-F_X(12)$ the probability that the first defective circuit made is _at least_ the thirteenth. Where $F_X(x)$ is the Cumulative Distribution Function (CDF).