-2
$\begingroup$

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?

1 Answers 1

2

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).

  • 0
    Sorry for so many editions and comments deleted, I do not know what is happening. But thank you very much for the answer. – 2017-02-07