Bayes' Theorem - any trick to figuring out what your E and F are?

Question

I'm having a lot of trouble figuring out what my E and F are supposed to be in Bayes' Theorem problems. Are there any tricks to this? Here's a really hard one for example...

A space ship communicates to base in a far away galaxy using bit strings. Suppose that in its transmissions it sends a 1 one-third of the time and a 0 two-thirds of the time. When a 0 is sent, the probability that it is received correctly is 0.9, and the probability that it is received incorrectly (as a 1) is 0.1. When a 1 is sent, the probability that it is received correctly is 0.8, and the probability that it is received incorrectly (as a 0) is 0.2.

Use Bayes’ theorem to find the probability that a 0 was transmitted, given that a 0 was received.

What would be my E and F in this example question?

Answer 1

You have two sets of events: either a $1$ or a $0$ was received, and either a $1$ or a $0$ was sent.

Let $E$ represent the event of a $0$ being sent in the sender's sample space, and let $F$ represent the event of a $0$ being read in the receiver's sample space.

In this case, you have $P(F | E) = 0.9$. This says "given that a $0$ was sent, the probability that I read a $0$ is 90%".

Answer 2

Let $E$ be the event a $0$ was transmitted, and $F$ the event a $0$ was received. We want $\Pr(E|F)$. I will not use the Bayes Theorem explicitly. You should not have any trouble recasting things slightly into the form you prefer. By the definition of conditional probability, we have $$\Pr(E|F)=\frac{\Pr(E\cap F)}{\Pr(F)}.\tag{1}$$ We calculate the two probabilities on the right-hand side of Equation (1).

First we find the harder one, $\Pr(F)$. We can receive a $0$ in two disjoint ways: (i) A $0$ was sent, and a $0$ was received or (ii) a $1$ was sent and a $0$ was received.

To find the probability of (i), note that a $0$ is sent with probability $\frac{2}{3}$. Given that a $0$ was sent, the probability a $0$ was received is $0.9$. It follows that the probability of (i) is $(2/3)(0.9)$.

In the language that you are probably looking for, we computed $\Pr(E)\Pr(F|E)$.

In the same way, we find that the probability of (ii) is $(1/3)(0.2)$.

Thus $\Pr(F)=(2/3)(0.9)+(1/3)(0.2)$.

Now we compute $\Pr(E\cap F)$. But we have already computed it, it is the probability of (i).

Finally, do the division that Equation (1) asks for.