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....


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:

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

$\frac{D1}{50!} chance

Then the $12^{th}$...:

$$C(10,9) \cdot C(40,2) \cdot (C1,1) = D2$$


Appears pattern of:

$$C(10,9) \cdot C(1,1) \cdot C(40,n-10) = Dn$$ where $n>10$


I'm quite sure this is incorrect. Please help.

  • 1
    $\begingroup$ It would be nice, if you would improve your question by using $\LaTeX$. I know, that you can do it. $\endgroup$ – callculus Aug 6 '15 at 16:04
  • 1
    $\begingroup$ For part a) I think the best way to go about it is to calculate the probability of finding $0$ broken or $1$ broken. Then subtract from $1$ to get the probability of at least $2$ broken. $\endgroup$ – Colm Bhandal Aug 6 '15 at 16:05
  • $\begingroup$ Sorry calculus. I dont know how to make fancy writings. $\endgroup$ – LinearSS Aug 6 '15 at 16:07
  • $\begingroup$ Colm, In part a) I tried to do what you say. Are my calculations incorrect? It gave me an answer of ~65%. I thought that was close to what I expected, assumed it to be correct. $\endgroup$ – LinearSS Aug 6 '15 at 16:10
  • $\begingroup$ I do not understand b). We are presumably inspecting one at a time. Is the question asking whether the $n$-th in the sample of $10$ is the last broken in the sample? Or are we testing them all, and asking for the probability that the $n$-th tested is the last broken? $\endgroup$ – André Nicolas Aug 6 '15 at 16:12

Your part (a) looks good to me. Your part (b) has a logical mistake.

When you say max ways is 50!, you're assuming that all pens are different and basically "arranging" them in different orders. If this is the case, then you should also apply arrangement in your D1, D2,... as well.

Just moving forwards with the logic you applied in D1, the final answer should be

Probability = [C(n-1,9) x 10! x 40!]/50!


C(n-1,9) -> Choose 9 places out of (n-1) places to place the broken pens. [nth place is taken by the 10th broken pen, and the other 9 should come before that]

10! -> Arranging those 10 broken pens

40! -> Arranging the other pens

50! -> Total ways of arranging

  • $\begingroup$ Okay, I see. The total arrangements = P(50;40,10) = 50!/(40!10!).......after arranging the 9 broken pens in the (n-1) postions = C(n-1,9), arrange the (n-10) working pens in C(n-10,n-10) = 1 way. Then divide by total arrangements, giving your probablity formula. Many Thanks $\endgroup$ – LinearSS Aug 6 '15 at 17:40

Part (a) looks good to me.

You are making a mistake in part (b). Note that when you say there are $50!$ ways of choosing 50 pens, you are thinking of each pen as unique. Among these $50!$ sequences, you would need to think of all the different ways to choose the 10 broken pens first, which is $10!$, then multiply by all the ways to choose the following 40 pens, which is $40!$. This gives all sequences in which you draw all 10 pens first.

Hint for (b): instead of using counting principles, try using probabilities. For $n = 10$, you need to choose all 10 broken pens to start. This probability is $$\frac{10}{50} \times \frac{9}{49} \times \dots \times \frac{1}{41}.$$

Do you see why? Then for $n = 11$, you know the $11^{th}$ draw has to be the $10^{th}$ broken pen. So you need to figure out the probability of getting 9 broken pens in the first 10 draws. And so on for the rest of the values of n.


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