How to find expected value of X and Y?

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?

• Compute the probability, $\psi$, that a given unit is defective. Then $E[Y]=\psi\times 100$. – lulu Nov 10 '19 at 12:02
• @lulu so it is n*p and what about X? I did it correctly? – Semyon Yurchenko Nov 10 '19 at 12:33
• It wasn't clear to me what you meant by $p$. If you meant what I called $\psi$, then yes. – lulu Nov 10 '19 at 12:51
• @lulu Thank you, very much! – 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:$$\mathbb EY=\sum_{i=1}^{100}\mathbb EY_i=100\mathbb EY_1=100P(Y_1=1)$$

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