Understanding different definitions of bayes theorem

Question

I am taking course on probability and reading about bayes theorem.

1. In Sheldon Ross' book, it given as $$P(E) = P(E|F)P(F) + P(E|F^C)P(F^C)$$ with note: Equation above states that the probability of the event E is a weighted average of the conditional probability of E given that F has occurred and the conditional probability of E given that F has not occurred – each conditional probability being given as much weight as the event on which it is conditioned had of occurring.

2. On wikipedia, it is given as $$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$

3. In Kenneth Rosen's book it is given as $$P(F|E)=\frac{P(F)P(E|F)}{P(E|F)P(F)\cup P(E|F^C)P(F^C)}$$

Though I understand proofs of all of those, I dont understand the connection between them or to say, how they all communicate the same meaning? Or am mistaking and they are talking about different stuff?

Answer 1

Equation 1. is known as the law of total probability

Equation 2. is Bayes' theorem.

Equation 3. is Bayes' theorem with the law of total probability applied on A and B on the $P(B)$ in the denominator.

Answer 2

Wikipedia's version is correct, $P(A|B)=\frac{P(B|A)P(A)}{P(B)}$.

However often you know $P(B|A)$, and you have to calculate $P(B)$. This is basically (1) in your expressions above, using that you can write

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^C)P(A^C)}$ which gives (3) - almost always your problems will require this form to solve real life problems.

So (2) and (3) are Bayes Theorem, (1) is not. And almost everytime you will end up using (3).

---

Remembering a typical example,

Example: One box has 3 white, 5 red. Another has 4 white, 4 red. I pulled out a box randomly, and a ball randomly from it. The ball is red. What's the probability it came from first box?

$A=$first box

$B=$red

Using (3) we get

$P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^C)P(A^C)}=\frac{5/8\times1/2}{5/8\times1/2+4/8\times1/2}=5/9$