Solve only by using the axioms of probability.

Question

An insurance company assumes each of its clients falls into one of three categories for risk: low, medium, and high. If clients are twice as likely to be low risk than medium risk, and three times as likely to be medium risk than high risk, what is the probability that a randomly selected client is high risk? Use only the axioms of probability.

So we have 3 options, $S=\{L,M,H\}$

Then $P(S)=(\{L\} \cup \{M\} \cup \{H\}$ Since the low risk is twice as likely we have $\frac{2}{4}$.

Since medium is three times as likely we have $\frac{3}{6}$. I'm not sure if I did that correctly. From here I have no idea where to go.

Answer 1

Suppose that $\Pr(L)=\ell$ is the probability of being low risk. Similarly, $\Pr(M)=m$ the probability of being medium risk and $\Pr(H)=h$ is the probability of being high risk.

Since these are the only three possibilities and a randomly selected client can only be one of these at a time, what does that imply?

That $\ell+m+h = 1$

We are told that clients are twice as likely to be low risk than medium risk. What does that imply about the relationship between $\ell$ and $m$?

That $\ell = 2m$

We are told that clients are three times as likely to be medium risk than high risk. What does that imply about the relationship between $m$ and $h$?

That $m = 3h$

Using these three observations, this should now be enough information now to solve for $h$

$\ell + m+ h = 1 \implies 6h + 3h + h = 1\implies \dots$