Probability when considering multiple independence I'm stuck on the following question and was wondering if anyone could give me some advice:
"In October 1994 a flaw in a Pentium chip was discovered that could result in a wrong answer when performing division. The manufacturer initially claimed that the chance of any particular division being incorrect was only 1 in 9 billion. Assuming that the 1 in 9 billion figure is correct and that results of different divisions are independent of one another, what is the probability that at least one error occurs in one billion divisions with this chip?"
The following is what I currently have, but I'm not sure if this is right and would appreciate if someone can tell me if I'm doing this incorrectly and how to approach the problem if so. 
P(one division) = 1/9,000,000,000
P(one error in 1 billion divisions) = 1,000,000,000 * (1/9,000,000,000) = 0.11

EDIT:
P(at least one error in 1 billion runs) = 1 - [(9 bill - 1)/9 bill]^1bill
P(no errors) = [(9 bill - 1)/9 bill]^1bill

 A: Hint: an analogous problem is what is the chance of rolling at least one 6 on 10 dice rolls? This is 1 $-$ the chance of rolling no 6s on 10 dice rolls = $1-(5/6)^{10}$. Taking $5/6$ to the tenth power is because each dice roll is independent and with independence the probabilities multiply.
Adopting your logic on the dice roll problem, the chance of rolling at least one 6 in 10 dice rolls would be $10\times 1/6 = 10/6 > 1$, and you can't have a probability $>1$. In your case, think about what your approach would lead to if the question were for 10 bilion divisions.
You can multiply probabilities of events when they are independent. You can add probabilities of events when they are mutually exclusive. But rolling a 6 on one dice roll isn't exclusive with rolling a 6 on another, so you can't add these probabilities (you'd overcount the case where you rolled two 6s). Similarly your approach of adding the probability of an error on one division with an error on another is incorrect because the errors are not mutually exclusive (you overcount the case where more than one error occurs).
A: 
P(no errors in 1 billion runs) = 1 - [(9 bill - 1)/9 bill]^1bill
P(at least one error) = 1 - answer above

As you previously mentioned, [(9 bill - 1)/9 bill]^1 bill is the chance that there is no failure in one billion divisions. One minus that is the probability of at least one failure, exactly like in the case of rolling the die. Why subtract that result from one again? 
