Probability - determine the probability for an event I have this probability question from homework
A system consists of $N$ chips in a parallel way, such that if at least one of the chips are working the system fully operates.
The probability that throughout a work day, a chip will get broken is $\frac{1}{3}$
Note that each of the events, where a chip is broken is an independent event.
At the end of a work day, the system still operates.
What is the probability that chip number 1 is operating (not broken)?
I struggle with understanding how I should approach this, because it seems at first sight that if the events are independent and we have the probability that each chip will be broken, then we could calculate the complement and get the probability that a chip still operates at the end of the day $(1-\frac{1}{3}) = \frac{2}{3}$
But I know it's wrong because I didn't consider all the other events in my calculation which I find hard to do.
Can I get some help how should I approach this?
Thanks.
 A: You are looking to calculate $\mathsf P(X_1\mid X\neq 0)$ when $X$ is the count of chips that work at the end of the day, given a rate of failure, $q=1/3$ (i.i.d. for each chip) and $X_1$ is the event that chip one works.   It is the probability that a specific chip works when given that at least one chip does.
Your approach needs to determine what kind of probability distribution $X$ has (and hence its probability mass function), then use the definition of conditional probability.
A: You want the conditional probability $\textsf P(\text{chip $1$ works}\mid\text{some chip works})$. This is
\begin{align}
\textsf P(\text{chip $1$ works}\mid\text{some chip works})&=\frac{\textsf P(\text{chip $1$ works}\cap\text{some chip works})}{\textsf P(\text{some chip works})}\\
&=\frac{\textsf P(\text{chip $1$ works})}{\textsf P(\text{some chip works})}
\\
&=\frac{\frac23}{1-\left(\frac13\right)^N}\;.
\end{align}
A: Hint: use Bayes theorem which is mathematically given by: $$P(E_i\mid A)=\frac{P(A\cap E_i)}{P(A)}$$ where $E_i$ is one of the event occured from $E_1,E_2,...,E_n$ so can you find it now. 
$E_1$ here is: chip $1$ is working.
A: Your argument is wrong, because knowing that the system is still operating at the end of the day allows you to exclude some of the possible cases. Here, it allows you to exclude the case that all of the chips are broken. So the true answer probability will be slightly bigger than $\frac{2}{3}$, because you have fewer possible outcomes.
One way to solve this is to apply Bayes' theorem
$$ P(A \mid B) = P(B \mid A) \frac{P(A)}{P(B)}.$$
where
$$A = \{\text{The first chip is still operating at the end of the day}\}$$
$$B = \{\text{The system is still operating at the end of the day}\}$$
