Question
A box consists of 11 components, 7 of which are defective.
(a) Components are selected and tested one at a time, without replacement, until a non-defective component is found. Let X be the number of tests required. Find P(X = 4).
(b) Components are selected and tested, one at a time without replacement, until two consecutive non defective components are obtained. Let X be the number of tests required. Find P(X = 5).
Answer
Seems to be a simple application of hypergeometric distribution! (let d = defective, n = normal)
For part a), if we have the following:
$ 4 $ normal components, $ 7 $ defective components. The probability of getting a normal component on the 4th test is: (i.e. {dddn})
$ P(X=4)=\frac{{7 \choose 3}{4 \choose 1}}{{11 \choose 4}} $
Part b) is similar, except we're aiming to get 2 defective components consecutively. This means {nnndd}
$ 4 $ normal components, $ 7 $ defective components. The probability of getting a normal component on the 4th test is:
$ P(X=5)=\frac{{7 \choose 3}{4 \choose 2}}{{11 \choose 5}} $
These answers aren't correct though, where is the flaw?
I've tried attempting this from the permutations perspective as well (i.e. $ 7 $ d's and and $ 4 $ n's, find P({dddn}) for part a and P({dddnn}) for part b.)
Did something similar for part b), to no avail.