elPatron434 wrote:
Hi
GMATinsight , I'm getting a different value of p. Could you please help me with where I'm going wrong?
2C1*p(1-p) + p^2 > 0.5
2p - 2p^2 + p^2 > 0.5
2p - p^2 - 0.5 > 0
4p - 2p^2 -1 > 0
2p^2 -4p +1 < 0
b^2 - 4ac = 16-8 = \sqrt{8}
Roots are 1+0.5\sqrt{2} or 1-0.5\sqrt{2}
So p< 1-0.5\sqrt{2}
I don't think that's the best way to approach the problem, but your work is entirely correct, except for the very last line. You have this inequality
2p^2 -4p +1 < 0
You found the two roots -- just for convenience, let's call them "r" and "s", where r = 1 + (√2/2), and s = 1 - (√2/2). So you can factor
(p - r)(p - s) < 0
That inequality means one factor is positive, the other is negative. Since r is bigger than s, p-r is smaller than p-s (when we subtract a larger value from p, we get a smaller result). So if one of p-r or p-s is negative, it must be p-r that is negative, since it's smaller than p-s. So p - r < 0, and p < r, and p - s > 0, and p > s. So the correct conclusion to draw from your quadratic inequality is the opposite of the one you reached -- it should be:
1 - (√2/2) < p < 1 + (√2/2)
Of course the righthand part of this inequality is useless, because we already know more than that; of course, p < 1. It's the lefthand part that is useful: 1 - (√2/2) < p, which, to one decimal place, means roughly that p > 0.3.
_________________
http://www.ianstewartgmat.com