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This probability is stumping ME!?
Light bulbs are shipped in lots of 24. The quality control department chooses a random sample of 8 bulbs from each lot and the lot will ship if there are no defectives found in the sample. What is the probability that the lots ships (i.e., she found no defectives in the sample) with 2 defective bulbs?(Presumably the defective bulbs are in the other 16).
There is no information given about the probability of a defective light bulb. Please enlighten me!
3 Answers
- blahb31Lv 61 decade agoFavorite Answer
We have a lot of 24 bulbs where 2 are defective and the other 22 are not.
If 8 bulbs are selected, then the number of light bulbs in the sample that are defective would have a hypergeometric distribution. (In the link below, N = 24, D = 2, n = 8.) We are looking for the probability that we get no defective light bulbs.
P(X = 0) = (2C0)(22C8)/(24C8) (nCr = n!/{r!(n-r)!})
= 319770/735471
= 0.43478 approximately
- 1 decade ago
im assuming that u mean there are 2 defective light bulbs in each box.
Sorry the 3 things make it difficult; 2 r easy. One useful function is the choose function{n!/[k!(n-k)!]}, which is the amount of ways that a person can choose n objects from k, so 8 choose 24 is the amounts of ways the department can pick 8 bulbs from 24. This has lots of plug and chug. im 13 and lazy so i dont want to hurt my brain on this.
Source(s): use wikipedia to look up the choose function - 1 decade ago
I'm not completely sure, its been awhile since I did probability, but I used my graphing calculator (TI-82) and I think I figured it out.
I entered: 22 nCr 8/ 24 nCr 8
Written:
C (22,8)
-----------
C(24,8)
because there are only 2 defective bulbs out of the lot of 24 bulbs It is assumed the other 22 are good.
22 nCr 8 because out of a possible 22 "good" bulbs 8 are picked
24 nCr 8 because there are 24 total bulbs to choose from and 8 of them need to be chosen.
I got .4348 (rounded) or 10/23
Sorry if I am wrong or misunderstood the question.
Good Luck!
