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**HallsofIvy** Suppose the percentage of left-handed people among mathematicians is n and among non-mathematicians is m with m< n. Suppose also that the percentage of mathematicians in the entire population is r.

I find it simplest, when working with percentages to assume some specific number- 100 is easiest for percentages! So suppose further that our population consists of 100 people. Then r of them are mathematicians and 1- r are not. rn of them will be "left-handed mathematicians" and r- rn= r(1-n) will be "right-handed mathematicians". m(1- r) will be "left handed non-mathematicians" and 1-r- m(1- r)= (1-m)(1-r) will be "right-handed non-mathematicians".

Now we have that the total number of left-handed people in the population is nr+ m(1- r) and the total number of right-handed people is r(1-n)+ (1-m)(1-r). The percentage of mathematicians among left-handed people is $\displaystyle \frac{nr}{nr+ m(1-r)}$ and the percentage of mathematicians among right-handed people is $\displaystyle \frac{r(1-n)}{r(1-n)+ (1-m)(1-r)}$

The problem, then, is to show that, if n> m, with m and n between 0 and 1, then $\displaystyle \frac{nr}{nr+ m(1-r)} > \frac{r(1-n)}{r(1-n)+ (1-m)(1-r)}$. Since all of those numbers are positive, I would begin by "cross multiplying".