Dang Probability!

Ail Bane · #1 Oct 19, 2007, 04:13 PM

I know this seems like an easy problem but I can't get it figured out.

The problem is:

If P(high) = .3, P(low) = .7,
P(favorable | high) = .9,
P(unfavorable | low) = .6,

then P(favorable) =

Normally I can figure stuff like this out, but usually I have more info than that. Well, there's always a first time.

I came up with .3, .27, and .1. I don't know what I'm doing wrong.

terryg752 · Oct 19, 2007, 05:09 PM

U = Unfav, F = Fav, H = High, L = Low

P(F/H) = .9

Hence P(U/H) = .1

[ Formula: P(A & B) = P(A) P(B/A) ]

P(U & H) = P(H) P(U/H) = .03

P(U & L) = P(L) P(U/L) = .42

P(U) = .03 + .42 = .45

HENCE: P(F) = 1 - .45 = .55

Ail Bane · Oct 19, 2007, 06:00 PM

Thanks for the help but does that mean that P(F) is .55?

terryg752 · Oct 19, 2007, 09:55 PM

Correct!

Ail Bane · Oct 19, 2007, 11:09 PM

Thanks, you're so helpful. Sorry I didn't see the .55 the first time.
I'm not sure if I could have done that on my own.

Its getting close to quiz time at the university so I needed it.

PS: I rated your answer.

terryg752 · Oct 20, 2007, 02:54 AM

Thanks, this was quite an interesting problem.

You are welcome.



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