Binomial probability on ports This problem appears very simple, but I am almost positive that it should not be so simple. 

10 ports. P1,P2,P3...P10 are connected to a computing device which polls
  them in order to check which is ready. 3 ports out of 10 are ready when
  we start the cycle. What is the probability that P1,P2 is not ready,
  and P3 is ready?

N,N,R,X,X,X,X,X,X,X
=0.5 x 0.5 x 0.5 x 1 x1 x 1 x 1 x 1 x 1 x 1
=0.125
Am I doing something wrong?
 A: You're not taking into consideration the fact that 3 ports are ready and 7 ports aren't ready, which means you have only $\binom{10}{3}$ permutations instead of $2^{10}$ permutations.
I believe it is called Dependent Probability (not sure about the English terminology for that).
In any case, you should ask yourself, out of all combinations of 3 ones and 7 zeros, how many start with 001. And since the remaining 7 digits consist of 2 ones and 5 zeros, the number of permutations that start with 001 is $\binom{7}{2}$.
Bottom line:
$\binom{7}{2}=\frac{7!}{2!5!}=21$
$\binom{10}{3}=\frac{10!}{3!7!}=120$
So the overall probability is $\frac{21}{120}=0.175$
A: We will assume that all combinations of three ports are equally likely to be the ones that are ready. The probability that Port 1 is not ready is $\frac{7}{10}$. Given that Port 1 is not ready, the probability that Port 2 is not ready is $\frac{6}{9}$, Given that 1 and 2 are not ready, the probability Port 3 is  ready is $\frac{3}{8}$. Multiply. 
