Thread: Probability Proof

x = 3p^2 − 2p^3

Show that x < p iff p < 1/2

note: p^2 = p'sqaured and p^3 = p'qubed

Can someone please tell me how to start off this problem?

thanks!

I probably should have just called it an algebraic proof however "p" in this case is a probability so I really should has said...

x = 3p^2 − 2p^3

Show that x < p iff 0 =< p < 1/2

In other words p can be from 0 to less than 0.5

Anyone?

First iff notation is a two way stree, we have to go forward as well as backward.

1st step to show x<p implies p<1/2.
note, x is a function of p, and x or 3p^2-2p^3 are equivalent.
for x<p, we can have 3p^2-2p^3<p
just do some rearrengement, you can get when p>1 or p<1/2 which allow
x<p, then we choose the p<1/2

2nd step, go backward, show p<1/2 implies x<p.
if we show x/p<1, then we showed x<p.
use (3p^2-2p^3)/p, can you show this is less than 1, when p<1/2?
of course you can.

still having problem, please point out.