Probability problem of Birth Rate and GDP Ronnie obtained data about the gross domestic product (GDP) and the birth rate for 170 countries.
He classified each GDP and each birth rate as either ‘low’, ‘medium’ or ‘high’. The table shows the
number of countries in each category.

One country is chosen at random from those countries which have a medium GDP and then a different
country is chosen at random from those which have a medium birth rate. Find the probability that both countries chosen have a medium GDP and a medium birth rate.
P.S. This is a question from Cambridge Advanced Subsidiary Level and Advanced Level
My Try
I tried this with a Tree Diagram. First set of branches as Medium GDP and Second Set of branches as Medium Birth Rate.This got me the answer $\frac{42}{74}\times\frac{42}{55}$. 
First I need to know if this is correct.Secondly 
can you help me to do this using conditional probability? Thanks a lot.
 A: Your answer is almost correct.
Overlooked in the calculation is that we are dealing with two different countries.
Let $B_1$ correspond with birth rate of firstly chosen country.
Let $G_2$ correspond with GDP of secondly chosen country.
Taking into account on forehand (so that no conditions for that are needed anymore) that the GDP of the first chosen country is "median" and the birth rate of the second chosen country is "median" we find:$$P(B_1=\text{median}\wedge G_2=\text{median})=P(B_1=\text{median})P(G_2=\text{median}\mid B_1=\text{median})=\frac{42}{74}\frac{41}{54}$$
Under condition "$B_1=\text{median}$" there are $55-1=54$ countries left that can be picked as second choice, and $42-1=41$ of them give "median" for GDP.
A: I'm not sure about the number $54$ you used calculate the answer.
I think it should be split into two cases for the denominator.

*

*Case $1$: first one chosen is medium GDP but not medium birth, in which case there are still $55$ ways to choose medium birth.
$(32)(55)$


*Case $2$: first one chosen is medium GDP and medium birth, in which case there are left with $54$ ways to choose medium birth.
$42(54)$
so my final answer should be $\frac{42(41)}{(32)(55)+42(54)}$
Let me know what you think?
