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...

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.
