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swissgirl

  • 3 years ago

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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  1. swissgirl
    • 3 years ago
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    @Libniz

  2. swissgirl
    • 3 years ago
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    How many ways can it be tested? Like how many samples are there?

  3. Libniz
    • 3 years ago
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    this sounds like geomterics variants

  4. Libniz
    • 3 years ago
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    P= probabilty of getting second defective p(1-p)^(n-1) something similiar to this

  5. swissgirl
    • 3 years ago
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    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

  6. Libniz
    • 3 years ago
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    4 C 2 ?

  7. Libniz
    • 3 years ago
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    4 C 2=6

  8. swissgirl
    • 3 years ago
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    Well no like the sample space would be the order that I test the components. idk probability is sooo confusing

  9. Libniz
    • 3 years ago
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    are you on deadline because I wanna take my time and think about it

  10. swissgirl
    • 3 years ago
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    nah i still have time. Gonna think abt it too. Like only if you have time

  11. swissgirl
    • 3 years ago
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    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

  12. kropot72
    • 3 years ago
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    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

  13. kropot72
    • 3 years ago
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    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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