would appreciate your help:
i have the following circuit 
eachone of the switches is closed in a probability of 0.7(and then the current is passing). each switch is independent of the other
and i need to calculate:
1)if switch 3 is open, what are the odds of a current passing from A to B?
2)if switch 3 is closed, what are the odds of a current passing from A to B?
my attempt:
1)if 3 is open, to pass current we need 1 &2 or 4 & 5. so is it 0.7 * 0.7 + 0.7*0.7-0.24?
2)if 3 is closed, to pass a current we need (1 or 4) and (2 or 5) = (0.7+0.7-0.49)*(0.7+0.7-0.49)?
are my calculations correct? please correct me if i have done something wrong.
thank you very much!
$1)$ is not correct because the chances of $1+2$ and $4+5$ are not disjoint. If you change the chance a switch is closed to $0.8$ you will get a result greater than $1$. Use $P(A \cup B)=P(A)+P(B)-P(A \cap B)$. You missed the last term.
For $2)$ you should not consider the chance that $3$ is closed because you are given that. You need $(1$ or $4)$ and $(2$ or $5)$