# Conditional Probability with Circuits

I am very sorry for the myriad of probability questions but I have a dam hard time trying to understand these questions:

This is the question:

Consider the following portion of an electric circuit with three relays. Current will flow from point a to point b if there is at least one closed path when the relays are activated. The relays may malfunction and not close when activated. Suppose that the relays act independently of one another and close properly when activated, with a probability of .9.

The question is: What is the probability that current will flow when the relays are activated?

1 - (0.1)^3

I don't understand why you are multiplying 0.1 * 0.1 * 0.1 and not adding 0.1 + 0.1 + 0.1. Since if any one of the three are open then the current will flow so isn't this technically a union which means you add?

• It's very simple: the multiplication principle. The relays are independent of each other. – Parcly Taxel Oct 12 '16 at 4:26
• @ParclyTaxel Oh ok. Thanks for pointing that out. So if they are dependent of each other then you would add them? – CapturedTree Oct 12 '16 at 5:23

Yes , it is a union .   The current will flow when at least one of the relays will pass it through .

However , you do not simply add because the events are not disjoint .   This then requires using the Principle of Inclusion and Exclusion (PIE) to avoid over-counting common outcomes .

\begin{align}\mathsf P(E_1\cup E_2\cup E_3) ~=~& {\mathsf P(E_1)+\mathsf P(E_2)+\mathsf P(E_3) \\[1ex]-\mathsf P(E_1\cap E_2)-\mathsf P(E_1\cap E_3)-\mathsf P(E_2\cap E_3)\\+\mathsf P(E_1\cap E_2\cap E_3)} \\[1ex]=~& 3(0.9)-3(0.9)^2+(0.9)^3\\=~& 0.999\end{align}

Since the events are independent , then the probabilities of the unions are the product of the probabilities of the events.

However the answer is somewhat easier to obtain by using complements .   The current will not flow only when all of the relays block it .

\begin{align}\mathsf P(E_1\cup E_2\cup E_3) ~=~& 1-\mathsf P(E_1^\complement\cap E_2^\complement\cap E_3^\complement) &~&\text{de Morgan's Rules}\\[1ex] =~& 1-\mathsf P(E_1^\complement)\,\mathsf P(E_2^\complement)\,\mathsf P( E_3^\complement) && \text{Independence} \\=~& 1-(1-0.9)^3 \\[1ex]=~& 0.999 \end{align}

That is all.

$\blacksquare$

• Thank you for the explanation! – CapturedTree Oct 12 '16 at 5:25

Current will not flow if $S_1 = S_2 = S_3 = 0$ which happens with probability $(1-0.9)^3$, hence current will flow with probability $1-(1-0.9)^3$.