## swissgirl 2 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.

• This Question is Open
1. swissgirl

@Libniz

2. swissgirl

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

3. Libniz

this sounds like geomterics variants

4. Libniz

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

5. 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

6. Libniz

4 C 2 ?

7. Libniz

4 C 2=6

8. swissgirl

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

9. Libniz

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

10. swissgirl

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

11. 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

12. 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

13. 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: $P(Y=4)=\frac{1}{2}\times 1=\frac{1}{2}$