# Broken Pens: combinations and probabilities

A container hold 50 pens. Exactly 10 pens are broken. What is the chance of finding:

a) In a random sample of 10 drawn from the container, 2 or more are broken?

b) The last broken pen to be the $n^{th}$ inspected?

Okay part (a) I think I've done correctly.

All the possible ways to draw 10 pens from a sample of 50:

$$C(50,10) = j$$

All the ways to select only working pens:

$$C(40,10) = k$$

All the ways to select exactly 1 broken pen:

$$C(40,9)*C(10,1) = l$$

Then all the ways to select at least 2 broken pens:

$$j - (k + l) = T$$

Then probability of finding at least 2 in sample of 10:

$$(T/j)*100 = P\%$$

Now I don't really understand the question in (b).....

Surely the smallest value of $n$ is $10$.

One cannot inspect the last broken pen before the other 9. Therefore only one way for $n=10, C(10,10)$

Now all the possible ways to inspect the 50 pens..... one by one....

50!

So then $\frac{1}{50!}$ chance for it to be the $10^{th}$

Now for the $11^{th}$ inspection to be the last broken pen:

$$C(10,9) \cdot C(40,1) \cdot C(1,1) = D1$$