What kind of distribution is this and how do I calculate the expected value Jack buys four items from a firm; the four are randomly selected from a large lot known to contain 10% defectives. 
Let $Y$ denote the number of defectives among the four that Jack has bought. 
Suppose that Jack will return the defectives for repair, and the repair is given by
$C=3Y^2 + Y + 2$
Find the expected repair cost. (Answer: 3.96)
So for $Y$ I was a bit confused about what type of distribution this follows. There is a large lot and 4 tyres are randomly selected. It looks like a negative binomial distribution to me so how would I calculate the expected value without k (where k represents the number of defective items).
$E[Y] = \frac{k}{0.1}$
Also. How do you deal with the $X^2$ in the repair cost formula. I know that for the linear part it will just be E[Y] + 2
 A: The distribution is binomial (characterized by a certain number of events occurring independently and with the same probability) - as the lot is large (and more importantly, the size is unspecified), you may assume that the probability does not change for each subsequent item.
A question which requires a negative binomial distribution could be "Jack keeps buying items until he receives 4 with defects. What is the probability that he bought at least __ items?".
$E(C) = 3E(Y^2) + E(Y) + 2$
Regarding the $Y^2$ term:
Recall that variance is defined: $Var(X) = E(X^2) - E(X)^2$. Hence, for the binomial distribution ($E(X) = np, Var(X) = np(1-p)$): $E(X^2) = Var(X) - E(X)^2 = np(1-p) + n^2p^2 = np(np-p+1)$
Hopefully this answers your question! Please don't hesitate to post below if you require further clarification.
A: The distribution is actually Binomial with paramemeters $n=4$ and $p=0.1$ (because we are dealing with a large lot, we assume a constant probability).
Your exercise then reduces to taking expectations of $C$. Let me know if you have any problems there now that the distribution has been specified.
A: The distribution is binomial: we take $n=4$ samples with probability $p = 0.1$ of defectiveness.
So the probabilities for $Y = 0,1,2,3,4$ are $(0.9)^4, 4\cdot(0.9)^3\cdot(0.1)^1,6\cdot(0.9)^2\cdot(0.1)^2, 4\cdot(0.9)^1\cdot(0.1)^3,(0.1)^4$ resp. using the formula you ought to know.
You can also compute the $C$ values for all of these $Y$,  filling in $Y=0,\ldots,4$ in your formula. Now multiply the values with the probabilities and add.
