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So far I have seen the following formulas:

Formula 1:

$P(E \vert F) = \frac{P(EF)}{P(F)}$

Formula 2:

$P(EF)= P(F)P(E \vert F)$

Formula 3:

$P(E) = P(E \vert F)P(F) + P(E \vert F^c)P(F^c)$

Now in the following example something happens that just doesn't make sense to me nor is it explained:

Suppose that a new policy holder has an accident with in a year of purchasing a policy. What is the probability that they are accident prone?

Previously in part one we found out that the event that a new policy holder has an accident with in a year has a .26 probability.

Let $A_1$ denote the event that the policy holder will have an accident with in a year of purchasing the policy and let A denote the event that the policy holder is accident prone.

Solution:

$P(A \vert A_1) = \frac{P(AA_1)}{P(A_1)}$

The following line is where I loose it and don't understand why A and $A_1$ flip:

$\frac{P(A)P(A_1 \vert A)}{P(A_1)}$

according to formula two it should have been:

$P(AA_1)= P(A_1)P(A \vert A_1)$

What are the scenarios when they flip and WHY?

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    In formula 2, you can interchange $E$ and $F$ to see that $P(EF) = P(E) P(F \mid E)$. Another way to look at it is to first note that $P(A A_i) = P(A_i A)$, then apply your formula 2. – 2017-01-19

2 Answers 2

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$P(AA_1)$ is the probability that both $A$ and $A_1$ hold. This is symmetric, so $P(A A_1) = P(A_1A).$ Thus we can apply the formula for conditional probability in two different ways, either conditioning on $A$ or $A_1$, giving $$ P(A A_1) = \frac{P(A|A_1)}{P(A_1)} = \frac{P(A_1|A)}{P(A)}.$$

This second equality can be rewritten as $$ P(A|A_1) = \frac{P(A_1|A)P(A)}{P(A_1)}$$ or $$ P(A_1|A) = \frac{P(A|A_1)P(A_1)}{P(A)}.$$

This is called Bayes' formula. It's very important and gets used a lot. Particularly in problems like these.

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Maybe this is done in those case where we are unable to find $P(A \cap A_1)$ easily.

And given $P(A_1 \vert A)$