What is the probability/percentage of $X$ people that $Y$? [bayes/law of total probability] Attempting the following problem, and it's going nowhere.
Suppose that $75$% of all people with credit records, improve their credit records within three years. Suppose that $18$% of the population at large have poor credit records, and of these only $30$% will improve their credit records within three years. What percentage of the people who will improve their credit records, are the ones who currently have good credit records?
My attempt has been (to find $P(G|I)$):
$R$: has a credit record, $B$: has poor credit record, $G$: has good credit record, $I$: will improve credit record.
$$P(G|I) = \frac{P(I|G)P(G)}{P(I)}$$
I end up proceeding with some manipulations but then being stuck with an expression for $P(G|I)$ in terms of $P(I)$, and then can't find a proper law of total probability expression that gets me a number for $P(I)$.
That is, the eventual point I get to is $$P(G|I|) = 1-\frac{P(I|B)P(B)}{P(I)}
= 1-\frac{0.054}{P(I)}$$ which is where everything further leads in circles.. any help would be much appreciated.
 A: If you make a Venn diagram of the current situation, you can visualise exactly what you need to find.

Let ABCD represent your total population - within this, the quadilateral EFGH depicts people with credit records. Now within these people, the quadliateral EIJF represents those who will improve in 3 years, and the circle represents those with poor credit scores
Now, to translate your question to this diagram, you need to find the proportion of people sitting in (EIJF - K) to (EHGF-K)
$$EIJF = 0.75 \cdot EHGF$$
$$K \cap EIJF = 0.3 K $$
$$P = \frac{(0.75)\cdot n(EHGF) - (0.3)\cdot n(K)}{n(EHGF)-n(K)}$$
Let the proportion of people having credit records at all be $p$. This gives us the missing link
$$P = \frac{(0.75)\cdot p\cdot n(ABCD) - (0.3)\cdot 0.18 \cdot n(ABCD)}{p \cdot n(ABCD)-0.18 \cdot n(ABCD)}$$
$$P = \frac{(0.75)\cdot p - (0.3)\cdot 0.18}{p -0.18}$$
Now as far as I can see, the question does not give us enough data to solve for $p$, assuming that everyone has a credit record, the probability is
$$P(p=1) = 0.848$$
Is this the answer given?
