A group of four components is known to contain two defectives. An inspector tests the components one at a time until teh 2 defectives are located. Once she locates the 2 defectives she stops testing but the second defective is tested to ensure accuracy. Let Y denote the number of the test on which the second defective is found. Find the probability distribution for Y.

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- swissgirl

How many ways can it be tested? Like how many samples are there?

- anonymous

this sounds like geomterics variants

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## More answers

- anonymous

P= probabilty of getting second defective
p(1-p)^(n-1)
something similiar to this

- swissgirl

Ok i see. But I would like to know how many ways it can be tested. like forget about the variances. I just would like to know how many sample spaces there are in this experiment

- anonymous

4 C 2 ?

- anonymous

4 C 2=6

- swissgirl

Well no like the sample space would be the order that I test the components. idk probability is sooo confusing

- anonymous

are you on deadline because I wanna take my time and think about it

- swissgirl

nah i still have time. Gonna think abt it too. Like only if you have time

- swissgirl

P(Y=2)=1/6
p(Y=3)=2/6
P(Y=4)=1/2
This is the answer in the back of my book

- kropot72

The answer in the back of your book checks out to the following extent:
(a) By using the hypergeometric distribution it is found that the probability of finding the two defectives in (Y=1) and (Y=2) = 1/6
(b) The probability that tests (Y=3) and (Y=4) will find the two defectives is therefore found from 1 - 1/6 = 5/6.
If the book answers for P(Y=3) and P(Y=4) are added we get 2/6 + 1/2 = 5/6

- kropot72

From the hypergeometric distribution:
\[P(Y=2)=\frac{\left(\begin{matrix}2 \\ 2\end{matrix}\right)\left(\begin{matrix}2 \\ 0\end{matrix}\right)}{\left(\begin{matrix}4 \\ 2\end{matrix}\right)}=\frac{1}{6}\]
The probability that one defective will be found when tests Y=1 and Y=2 are performed is found from:
\[P(1defectiveInY1andY2)=\frac{\left(\begin{matrix}2 \\ 1\end{matrix}\right)\left(\begin{matrix}2 \\ 1\end{matrix}\right)}{\left(\begin{matrix}4 \\ 2\end{matrix}\right)}=\frac{4}{6}\]
The probability that the second defective will be found on the third test given that the first defective has been found on Y=1 and Y=2 is 1/2.
The combined probability of finding the first defective on tests Y=1 and Y=2 and then finding the second defective on test Y=3 is found from:
\[P(Y=3)=\frac{4}{6}\times \frac{1}{2}=\frac{2}{6}\]
The probability that one defective will be found when tests Y=1, Y=2 and Y=3 are performed is found from:
\[P(1defectiveInY1Y2Y3)=\frac{\left(\begin{matrix}2 \\ 1\end{matrix}\right)\left(\begin{matrix}2 \\ 2\end{matrix}\right)}{\left(\begin{matrix}4 \\ 3\end{matrix}\right)}=\frac{1}{2}\]
Given that one defective was found in tests Y=1, Y=2 and Y=3 the probability of finding the second defective on test Y=4 is 1.
The combined probability of finding the first defective on tests Y=1, Y=2 and Y=3 and then finding the second defective on test Y=4 is found from:
\[P(Y=4)=\frac{1}{2}\times 1=\frac{1}{2}\]

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