My question is as follows:
A batch of 50 different automatic typewriters contains exactly 10 defective machines. What is the probability of finding:
(c) The first defective machine to be the kth machine taken apart for inspection in a random sequence of machines?
(d) The last defective machine to be the kth machine taken apart?
We were given the answer of $$\frac{P(40,k-1)*10}{P(50,k)}$$ for part c) which kind of makes sense to me, but the explanation we were given was kind of convoluted.
I am trying to solve part d) using the same method used for part c). I know it's a negative binomial situation, but that isn't taught in this course and we can only use combinatorics to solve this question.
I am unsure where to begin, but here is probably my best attempt, and I have no idea if I'm on the right track or not.
$$\frac{P(k-1,9)}{C(50,10)}$$ My reasoning behind this answer is that 9 of the 10 broken machines must be selected in the first k-1 machines (numerator), and then there is only 1 way for the 10th defective machine to be the kth machine. Then, the denominator is the total number of ways 10 broken machines could appear in a sequence of 50 machines.
Thank you and any help is appreciated.
In part d), there are $10$ ways to choose which defective machine will be the $k$th inspected. If there are $9$ defective machines among the first $k-1$, then there are $k-10$ good ones, and there are ${40\choose k-10}$ ways to choose them. There are ${50\choose k}$ ways to choose the first $k$ machines, so the probability is $${10{40\choose k-10}\over{50\choose k}}$$
In case you are unacquainted with the notation ${n\choose k}$ means the number of ways to choose $k$ items from $n$, the same thing you are writing as $C(n,k).$ Note that when $k<0,\ {n\choose k}=0$ so that the probability is $0$ when $k<10$, as it should be.
I might mention that you should see there is something wrong in your answer to part d), because you have permutations in the numerator and combinations in the denominator. Since we want to know what fraction of the combinations in the denominator meet certain requirements, we will have combinations in the numerator, too. Otherwise, you're comparing apples and oranges, as they say.
Finally, if you re-do part c) in terms of combinations rather than permutations, I think you'll find it more straightforward.