1. Inequality query

Hi,
I'd be grateful if anyone could elucidate why this

$\sum_{k=1}^2 \binom{2}{k} (1-p)^k p^{2-k} > \sum_{k=2}^4 \binom{4}{k} (1-p)^k p^{4-k}$

is equivalent to this

$\sum_{k=0}^1 \binom{4}{k} (1-p)^k p^{4-k}> \binom{2}{0} (1-p)^0 p^{2}$

I guess it should be obvious but I can't quite see why. Thanks in advance. MD

2. Re: Inequality query

Originally Posted by Mathsdog
Hi,
I'd be grateful if anyone could elucidate why this

$\sum_{k=1}^2 \binom{2}{k} (1-p)^k p^{2-k} > \sum_{k=2}^4 \binom{4}{k} (1-p)^k p^{4-k}$

is equivalent to this

$\sum_{k=0}^1 \binom{4}{k} (1-p)^k p^{4-k}> \binom{2}{0} (1-p)^0 p^{2}$

I guess it should be obvious but I can't quite see why. Thanks in advance. MD
Consider the binomial expansions of $((1-p)+p)^2$ and $((1-p)+p)^4$

CB