Probability problem : A lot of 100 bulbs from a manufacturing process is known to contain 10 defective and 90 non ... Problem : 
A lot of 100 bulbs from a manufacturing process is known to contain 10 defective and 90 non defective bulbs. If 8 bulbs are selected at random, what is the probability that there will be at least one defective bulb. 
Method I : 
Probability that no bulb will be defective , $P(X =0) $  
$$P(0) ={}^8C_0 p^0 q^8 =q^8 = (\frac{9}{10})^8$$ 
where q probability that a bulb selected is non defective $$\therefore q = 1-p = 1-\frac{1}{10} = \frac{9}{10}$$
and p ( probability of bulb drawn is a defective ) = $$\frac{10}{100} = \frac{1}{10}$$
Now probability that at least one bulb is defective = $$1 -P(0) = 1-(\frac{9}{10})^8$$
Method II :  
But I want to find the answer in another way,
Probability of drawing non defective bulb 
$$=  \frac{90}{100} \times \frac{89}{99} \times \frac{88}{98} \times...........\times \frac{82}{92} ..........(i)$$ 
Now the probability of drawing at least one bulb will be defective 
$$= 1-(i)  = 1-   \frac{90}{100} \times \frac{89}{99} \times \frac{88}{98} \times...........\times \frac{82}{92}$$
But this is not the correct answer ... please suggest the correction here..... thanks...
 A: The problem says: "A LOT of 100 bulbs from a manufacturing process is known to contain 10 defective and 90 non defective bulbs". It doesn't say that there is only 100 bulbs (10 defective and 90 non defective), it means that in every group of 100 bulbs should be 10 defective and 90 non defective bulbs by probability, in other words - the rate of defective bulbs is 10/100 and the rate of non defective bulbs is 90/100. So consider that you have an infinite number of bulbs which consists of 10% of defective and 90% of non defective bulbs. That's why the 1st solution gives you the right answer :) The problem could also say "A lot of 10 bulbs from a manufacturing process is known to contain 1 defective and 9 non defective bulbs", it doesn't matter on the numbers, but on the rate of bulb types.
A: Method II is the correct method. However you are drawing $8$ bulbs, not $9$, and so the answer becomes $$1 - \left(\frac{90}{100} \cdot \frac{89}{99} \cdot \ldots \cdot \frac{83}{93}\right)$$
A: This is a classic case of how conditional probability changes the independence of events, i.e. draws from the lot.
Method 2 assumes that you look at each bulb when you draw it, which means that you gain information about the remaining bulbs, and the next draw is now conditionally dependent on the previous draws (e.g. drawing 7 bad bulbs in a row changes the odds of drawing a bad bulb from 10/100 = 10% to 3/93 = 3.23%.  
The problem is asking for 8 bulbs selected at random.  This means that you either scooped out 8 bulbs at once or drew 8 bulbs individually with a blindfold on, and then took the blindfold off and looked at all 8 together.  This keeps the draws independent.  This is what the problem is asking.  Only Method 1 is correct; Method 2 answers a different question.  
