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swissgirl
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.
How many ways can it be tested? Like how many samples are there?
this sounds like geomterics variants
P= probabilty of getting second defective p(1-p)^(n-1) something similiar to this
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
Well no like the sample space would be the order that I test the components. idk probability is sooo confusing
are you on deadline because I wanna take my time and think about it
nah i still have time. Gonna think abt it too. Like only if you have time
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
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
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: \[P(Y=4)=\frac{1}{2}\times 1=\frac{1}{2}\]