Once 100 detailes were made 2 experts check them for defects. There are two types of defects that are independent. Probability of first defect is 0.02 and second is 0.05. What are the expected values of X-number of checked details until first defected detail comes up and Y-expected value of number of defected details in all 100 details. Firsly we can find X as expected value of geometrical distribution E=1/p. It equals 14.5. But how we can find Y? Is it counted like n*p? And am i right with X?

  • $\begingroup$ Compute the probability, $\psi$, that a given unit is defective. Then $E[Y]=\psi\times 100$. $\endgroup$ – lulu Nov 10 '19 at 12:02
  • $\begingroup$ @lulu so it is n*p and what about X? I did it correctly? $\endgroup$ – Semyon Yurchenko Nov 10 '19 at 12:33
  • $\begingroup$ It wasn't clear to me what you meant by $p$. If you meant what I called $\psi$, then yes. $\endgroup$ – lulu Nov 10 '19 at 12:51
  • $\begingroup$ @lulu Thank you, very much! $\endgroup$ – Semyon Yurchenko Nov 10 '19 at 14:41

I cannot find any flaw in your calculation and thinking.

The probability that a detail is defect is $1$ minus the probability that it is not defect, so equals: $$p:=1-(1-0.02)(1-0.05)=0.069$$

Your first calculation (applying geometric distribution) is okay and gives outcome: $$\frac1p=14.49275$$

Your idea for the second calculation is also okay and gives outcome: $$100\times0.069=6.9$$

For this observe that you can write the number of defectives as:$$Y:=\sum_{i=1}^{100}Y_i$$where $Y_i$ takes value $1$ if detail $i$ appears to be defect and takes value $0$ otherwise.

You can find the expectation of $Y$ by applying linearity of expectation and symmetry leading to:$$\mathbb EY=\sum_{i=1}^{100}\mathbb EY_i=100\mathbb EY_1=100P(Y_1=1)$$

  • $\begingroup$ Thank you, very much! $\endgroup$ – Semyon Yurchenko Nov 10 '19 at 14:41

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