Probability of an event in a chain of events with given conditional probabilities Meredith Watts, police chief of Shorewood, knows that the probability of any assault that occurs in the city being reported to the police is .5.  If an assault is reported, the probability that someone will be arrested is .6.  If someone is arrested, the probability of a conviction is .9.  If a person is convicted of assault the probability of actually serving jail time is .7.  Given that an assault occurs, what is the probability of a person serving jail time for the crime?



I knew that by the question stating "given" that an assault occurs that I need to use the rule of conditional probability. I know the formula for that is $Pr(A |B) = Pr (A \cap  B)/Pr(B)$. 

So I thought the best way to get started would be to use a probability tree, but I have never done that before and I am not exactly sure how to get started.
Hint:

Let (symbol) denote the event of (event)

As - Assault

R - Report

A - Arrested

C - Conviction

S - Serving

We know ( Note that | means given)

P(R|As) = 0.5

P(A|R) = 0.6

P(C|A) = 0.9

P(S|C) = 0.7

and

We want to know:

P(S|As) = ?

Before computing this mathematically, through conditional probability, let us think about this intuitively using probability tree diagrams.



anything with "^c" means complement as in the event did not occur e.g. R^c ($R^c$) means that the assault was not reported.

It seems that the only way to arrive at S from As is through R, A, C and S. This is correct as unreported assaults do not result in arrests, not being arrested cannot result in a conviction and not being convicted means not serving.

When going through the probability tree this way, we multiply the probabilities (eg the probability of getting double 6's when rolling a pair of fair dice is (1/6)*(1/6) = 1/36).

Thus, the answer would be 0.5*0.6*0.9*0.7, but how do we state this formally? More importantly, why does going through the probability tree in such a way call for multiplication?



Let us note that if we serve a sentence, it follows that we have been convicted.

If we were given that a conviction happened and were asked to compute the probability for a sentence, the answer would be 0.7, and we are done.

Is being convicted given? No. Go up the tree.

Being convicted means we were arrested. 

Is being arrested given? No. Go up the tree.

Being arrested means we were reported. Given? No. Up.

Being reported means we committed an assault. This one is given.

Oh also, note that $S = R \cap A \cap C \cap S$, that is the event S is equal to the events R, A, C and S all happening together since there is no way someone can be sentenced if they are not first convicted, arrested, reported and are guilty of assault.

Recall we noted earlier: It seems that the only way to arrive at S from As is through R, A, C and S. This is correct as unreported assaults do not result in arrests, not being arrested cannot result in a conviction and not being convicted means not serving.

Thus, $P(S | As) =  P(R \cap A \cap C \cap S | As)$.

By definition of conditional probability, we have

$P(R \cap A \cap C \cap S | As) = \frac{P(R \cap A \cap C \cap S \cap As)}{P(As)}$.

If we could say that the events R, A, C, S and As are independent, then we would have $\frac{P(R \cap A \cap C \cap S \cap As)}{P(As)} = \frac{P(R)P(A)P(C)P(S)P(As)}{P(As)}$.

Unfortunately, they are not independent, I think. If not, there is a rule here we can use called the chain rule of probability.

We compute $P(R \cap A \cap C \cap S \cap As)$ as follows:

$P(R \cap A \cap C \cap S \cap As) = P(S | C \cap A \cap R \cap As) P(C | A \cap R \cap As) P(A | R \cap As) P(R|As) P(As)$

Hey look something we know, P(R|As) = 0.5.

What about the others?

We could use the extremely long and tedious process of doing things like:

$P(A | R \cap As) = \frac{P(A \cap R \cap As)}{P(R \cap As)}$

or we could take note of the fact that

$R \cap As = R$

$A \cap R \cap As = A$

$C \cap A \cap R \cap As = C$

This is true by similar reasoning to $S = R \cap A \cap C \cap S$.

Thus, we have

$P(R \cap A \cap C \cap S \cap As) = P(S|C) P(C|A) P(A|R) P(R|As) P(As)$

Hey wait, we don't know P(As). All we know is that it happened.

That's okay, we don't want to compute $P(R \cap A \cap C \cap S \cap As)$. We instead want to compute

$P(R \cap A \cap C \cap S | As) = \frac{P(R \cap A \cap C \cap S \cap As)}{P(As)} = \frac{P(S|C) P(C|A) P(A|R) P(R|As) P(As)}{P(As)}$.



Alternatively, you can write $P(R \cap A \cap C \cap S \cap As| As)$ and then note that $P(As | As) = 1$.



To understand chain rule of probability more, you may want to check out The Birthday Problem. It's a popular probability problem.
Required probability

$=(0.5)(0.6)(0.9)(0.7)$

$=0.189$