I have a ratio between two probabilities $$ \frac{(1-p)^{1000}+1000*p*(1-p)^{999}}{ \binom{1000}{2}*p^2*(1-p)^{998}}$$ and it is asked to show that this ratio is >1 (without a calculator)
$p=2/1000$
I arrived to this passage. $$ \frac{(1-p)^{1000}+1000*p*(1-p)^{999}}{ \binom{1000}{2}*p^2*(1-p)^{998}}= \frac{(1-p)^{2}+1000*p*(1-p)}{ 500*999*p}$$ Then the suggested solution says $$ \frac{(1-p)^{2}+2*p*(1-p)}{ 999*p}$$ but i don't understant where does it come from? it there something that I'm ignoring?
Write ${1000 \choose 2}=\frac 12 \cdot 1000 \cdot 999$. Divide numerator and denominator by $(1-p)^{998}$ and use the extra factor of $p$ in the denominator to cancel the $\frac {1000}2$