1000 lamps: 10% of a stock of 1000 light bulbs are faulty....

lugligino

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10% of a stock of 1000 light bulbs are faulty. If you randomly choose a sample of 10 light bulbs, what is the probability that one is faulty?

I tried this:
There are ten possible events:
1) I choose the faulty bulb at the first attempt.
2) I choose the faulty bulb at the second attempt.
...
10) I choose the faulty bulb at the tenth attempt.

By definition, the probability P1 to immediately choose the first faulty bulb is 10%.
The P2 probability of encountering a defective bulb at the second choice is (900/1000) * (100 / (1000-1)).
The P3 probability of encountering a defective bulb at the third choice is (900/1000) * ((900-1) / (1000-1)) * (100 / (1000-2)).
...
The probability P10 to meet a defective light bulb at the tenth choice is (900/1000) * ((900-1) / (1000-1)) * ((900-2) / (1000-2)) * ... * ((900-8) / (1000-8)) * (100 / (1000-9))

Summing up the probabilities of the ten independent events obtained above I get a 65.3% chance that it seems to be too high. Where am I wrong?
 
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10% of a stock of 1000 light bulbs are faulty. If you randomly choose a sample of 10 light bulbs, what is the probability that one is faulty?

I tried this:
There are ten possible events:
1) I choose the faulty bulb at the first attempt.
2) I choose the faulty bulb at the second attempt.
...
10) I choose the faulty bulb at the tenth attempt.

By definition, the probability P1 to immediately choose the first faulty bulb is 10%.
The P2 probability of encountering a defective bulb at the second choice is (900/1000) * (100 / (1000-1)).
The P3 probability of encountering a defective bulb at the third choice is (900/1000) * ((900-1) / (1000-1)) * (100 / (1000-2)).
...
The probability P10 to meet a defective light bulb at the tenth choice is (900/1000) * ((900-1) / (1000-1)) * ((900-2) / (1000-2)) * ... * ((900-8) / (1000-8)) * (100 / (1000-9))

Summing up the probabilities of the ten independent events obtained above I get a 65.3% chance that it seems to be too high. Where am I wrong?

First find the p(d d' d' d' d' d' d' d' d' d') where d = defective and d' = not defective and where the 1st one is defective and 2nd through 10th not defective. Then multiply this probability by the number of ways to obtain one defective and 9 non defective.
 
Thanks a lot. I should have solved it with your hint:
p(d d' d' d' d' d' d' d' d' d') is equal to (1/10)^1*(1-1/10)^(10-1) and there a 10 ways to obtain it so the probability to have one defect in the lot is 0.9^9.
 
Consider this:
If you test a lamp and the 1st is defective the event has probability 100/1000 to happen, the 2nd must be not defective and the probability is 900/999, the 3rd 899/998, ..., the 10th 892/991.
If the second is defective 1st lamp is not defective and the event has probability 900/1000, the 2nd is defective and event has probability 100/999, the 3rd is not defective and event has probability 899/998, ..., the 10th 892/891.
You can consider the the failure at third, fourth, fifth, ... , tenth position and every time you have the same probability to have one defective lamp and it is 0.0389 that, multiplied by 10, is slightly different from 0.9^9=0.387.
 
Consider this:
If you test a lamp and the 1st is defective the event has probability 100/1000 to happen, the 2nd must be not defective and the probability is 900/999, the 3rd 899/998, ..., the 10th 892/991.
If the second is defective 1st lamp is not defective and the event has probability 900/1000, the 2nd is defective and event has probability 100/999, the 3rd is not defective and event has probability 899/998, ..., the 10th 892/891.
You can consider the the failure at third, fourth, fifth, ... , tenth position and every time you have the same probability to have one defective lamp and it is 0.0389 that, multiplied by 10, is slightly different from 0.9^9=0.387.
The answer is (100 choose 1)(900 choose 9)/(1000 choose 10) ~0.389.
 
Thanks a lot. I should have solved it with your hint:
p(d d' d' d' d' d' d' d' d' d') is equal to (1/10)^1*(1-1/10)^(10-1) and there a 10 ways to obtain it so the probability to have one defect in the lot is 0.9^9.
What you concluded is it with replacement or non replacement and which one should you be using?
 
The answer is (100 choose 1)(900 choose 9)/(1000 choose 10) ~0.389.
Your answer is nice and elegant. It can be easily extended to the case of 2, 3, ..., 10 defective lamps.
 
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Yes it can be easily extended to the case of 2, 3, ..., 10 defective lamps
I tried to understand the meaning of your formula:
Lamp stock (S pieces) is composed by D*S defective lamps and (1-D)*S non defective lamps (0<D<1).
A lot of N lamps is extracted from the stock. What is the probability that you have n defective lamps in the lot?
The stock is composed by D1, D2, …, DD*S defective lamps and N1, N2, …, N[(1-D)*S] not defective lamps.
The lot is composed by n defective lamps and (N-n) non defective lamps so you can choose the defective part of the lot in

(D*S)*(D*S-1)*(D*S-2)*…*(D*S-n+2)*(D*S-n+1)

ways. You have

n*(n-1)*…*2*1 = n!

different sequences of the same n chosen defective lamps. If you don’t regard about the order of the chosen lamps the number of way to choose the defective lamps is

(D*S)*(D*S-1)*(D*S-2)*…*(D*S-n+2)*(D*S-n+1)/n! = (D*S)!/[(D*S-n)!*n!] = Comb (D*S;n) (that is an Excel function, too).

In the same manner you can choose the not defective parts of the lot in Comb[(1-D)*S;N-n] ways.

You can combine defective (n items) and not defective (N-n items) parts to form the lot in

Comb(D*S;n)* Comb[(1-D)*S;N-n]

ways.
The probability to have n defective lamps is given by the ways above divided by the total number of N lamps lots you can choose in the S lamp stock and it is Comb (S;N) so the answer is

Comb(D*S;n)* Comb[(1-D)*S;N-n]/Comb(S;N)

that should be the generalization of your formula, if I did it right.
Do you agree?
 
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