<p>dravndn546,</p>
<p>I am not sure if you were addressing to me. If you were, I was actually chill. I was merely illustrating how the math should be. ;)</p>
<p>Professor101,
Actually, my analysis does include independent events as the equation apply regardless:
P(D or C) = P(D) + P(C) - P(D & C)
= P(D) + P(C) - P(D) * P(C given D)
= P(D) + P(C) - P(D) * P(C) ("C given D" = "C" if C and D are independent)</p>
<p>What I was saying is one needs to find out what P(D&C) is numerically. We know it can be written as P(D)*P(C given D). But what is P(C given D) then? It's not P(C) (independent) and its not 1 either (everyont that gets in Dartmouth gets in Cornell). It's somewhere in between (I think that's probably what you were thinking; in this case, it's probably closer to 1 however). I actually meant to say this but I apologize for not getting this across. And P (C or D) would therefore be greater than P(C) alone -- so I agree that applying more would increase your chance. There's also implicit assumption that P(C) and P(D)...remain constant but they may shrink as one applies to too many schools and gets sloppy for each application. So that's when applying more could hurt. So then we can ask: "what's the right balance?". :)</p>