# Question on negative binomial distribution

A shipment of 2500 car headlights contains 200 which are defective. You choose from this shipment without replacement until you have 18 which are not defective. Let $X$ be the number of defective headlights you obtain.

find the probability function f(x)

• If the sample is without replacement then you should be looking at the negative hypergeometric distribution – Henry Feb 5 '15 at 7:35
• @Henry Yes you are right, I do not understand the concept of combining both the methods- negative and hypergeometric. could you please elaborate. thanks for trying! – L887 Feb 5 '15 at 7:51

The process of choosing ends, the moment we pick $18^{th}$ non-defective car headlight.

Before picking $18^{th}$ non-defective car headlight, we have picked $17$ non-defective and x defective car headlights.

Think of this process or event as sequence of two steps.
1. Picking 17 non-defective car headlights.
2. Picking x defective car headlights.

The number of possible outcomes in the event is then product of combinations obtained in each of the step.
1. number of non-defective car headlights = 2500-200 = 2300
Pick 17 non-defective car headlights from 2300 non-defective ones $2300 \choose 17$
2. number of defective car headlights = 200
Pick x defective car headlights from 200 defective ones $200 \choose x$

Number of Possible Outcomes in Event = $2300 \choose 17$$200 \choose x Sample Space = 2500 \choose 17+x \therefore Pr(X=x) = {2300 \choose 17} {200 \choose x}/{2500 \choose 17+x} • your answer is different from the below answer which one is correct? – L887 Feb 5 '15 at 17:04 • Given answer is correct. I think in the below answer, attempt has been made to evaluate the resulting combinatorial expression. I have just kept it simple. – Curious Feb 6 '15 at 8:36 I would usually point you at the Wikipedia article on the negative hypergeometric distribution but in this case it is not helpful. The probability that there are X=x defectives by the time you find the 18th not-defective item is the probability that there are x defectives in the first x+17 items multiplied by the conditional probability that the next item is not defective. So$$\Pr(X=x)=\dfrac{\displaystyle{{2300 \choose 17} {{200} \choose {x}}}}{ \displaystyle{2500 \choose x+17}}\cdot \dfrac{2283}{2483-x}.$$• I didn't understand this point:" the probability that there are x defectives in the first x+17 " ? – L887 Feb 5 '15 at 8:39 • Before you find the$18$th non-defective, you need to find$17$non-defectives and$x\$ defectives. – Henry Feb 5 '15 at 8:52