I need help with a question that has confused me. picture of the system : A water flow system has 4 valves
For the water to flow A and D must be open and also at least one of the valves B and C. The probability that a particular valve will be open is 0.8 and not dependent on the other valves.
- What is the probability that water will flow in the system? my answer : The probability that at least one of the valves B and C will be open is : 0.8*0.8+0.8*0.2+0.2*0.8 = 0.96 the probability that water will flow in the system is : 0.8*0.96*0.8 = 0.6144
2.It is known that no water has flowed into the system, what is the probability that this was due to the fact that it was only a closed D valve?
And here I was really unsuccessful, I found some answers but I have no way to check which one is correct, anyone can help?
Your answer to the first question is correct. A simpler way to compute the probability that $B$ or $C$ is open is to compute the complementary probability; i.e., the probability that $B$ and $C$ are closed. This occurs with probability $(0.2)^2 = 0.04$, hence the probability that at least one of $B$ or $C$ is open is $1 - 0.04 = 0.96$.
For the second part, it is easier to conceptualize the $B$ and $C$ valves as a single valve, say $E$, whose probability of being open is $0.96$ as you computed in the first part. So now we have a series of valves, $A$, $E$, and then $D$. Given that at least one of them is closed, thus no flow, what is the set of outcomes in which opening $D$, if it was not already open, would result in flow? In other words, $A$ and $E$ must be open and $D$ is the only closed valve. Otherwise, if either $A$ or $E$ were closed, opening $D$ if it was not already open, would not result in flow. So, what is the conditional probability in this case? Specifically, what is $$\frac{\Pr[(A \text{ open}) \cap (E \text{ open}) \cap (D \text{ closed})]}{\Pr[\text{no flow}]}?$$ Note that the probability in the denominator is the complement of the probability you got in the first part of your question.