Heisenberg Electric Circuit

From the diagram, we see that there are two paths from left to right and both of these paths have three switches. The probability of one path working is $\frac{1}{27}$ , while the probability of both paths working is $\frac{1}{243}$ . By the Principle of Inclusion-Exclusion, the total probability is thus $\frac{1}{27}+\frac{1}{27} -\frac{1}{243} = \boxed{\frac{17}{243}}.$

Woah I just came up with a formula for adding up probabilities in parallel!

So we all know that if we had two of those switches in series (each with a probability of being closed of $\frac{1}{3}$ ), then the probability of both being closed is $\frac{1}{9}$ . To show this graphically, we can model it the probabilities with X 's and O 's.

$X O O = \frac{1}{3}$

$X O O = \frac{1}{3}$

Total probability $= \frac{1}{9}$

By drawing a little arrow connecting two symbols (either an X or an O ), and doing that for each possible combination, you can just count the number of connections you made that have two X's , divide that by the total number of connections you made, and that will give you the probability that both of the switches in series will be closed.

The only difference you make to find out the probability of switches in parallel (as they are in this problem) is, after doing the whole drawing of X 's and O 's, you count the number of connections you made that have even one X in them . Think about that if the underlying logic seems unclear.

After toying with that self-made illustration that I made, I started to see what 'net' probabilities I would get for various combinations of two different probabilities in parallel. To my surprise, I came up with this mathematical description.

Let $P_1$ be the probability of the first 'switch' in parallel being closed, and $P_2$ be the probability for the second. The net probability, $P_{net}$ (the probability that either or both will be closed), is:

$P_1 + P_2 - (P_1 \times P_2) = P_{net}$

Applying this to the original problem, we start off by reducing the two switch probabilities in each of the pathways to

$\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$ .

Then, using the formula that I just wrote down, we get the 'net' probability of either pathway being closed:

$\frac{1}{9} + \frac{1}{9} - (\frac{1}{27}) = \frac{17}{81}$

Finally, we just do another probability in series for the last switch on the right:

$\frac{17}{81} \times \frac{1}{3} = \boxed{\frac{17}{243}}$