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...

Method II is the correct method. However you are drawing $8$ bulbs, not $9$, and so the answer becomes $$1 - \left(\frac{90}{100} \cdot \frac{89}{99} \cdot \ldots \cdot \frac{83}{93}\right)$$
