Odds in Probability of Green Eyes

[Math Man900]

Homework Statement

Of 27 students in a class, 11 have blue eyes, 13 have brown eyes, 3 have green eyes. If 3 students are chosen at random what are the odds of the following occurring?

Only 1 has green eyes.

Homework Equations

Odds = P( s )/P( f ) = P( s )/( 1 - P( s ))
Probability success = P( s )
Probability fail = P( f )

The Attempt at a Solution

P( s ) = (3/27) * (24/26) * (23/25) = 92/975
P( f ) = 1 - P( s ) = 883/975
Therefore: Odds = 92/883
I went over this problems many many times, and i believe this is the correct answer, but the back of the book says the answer is 92/233. There is a chance the back of the book is wrong. any ideas? thanks a lot.

 

[Math Man900]

Nevermind. there are 3 ways to get 1 green person (1st,2nd, or 3rd) which means P( s ) = 3*92/975. and when you work out P( f ) you find the odds are: 92/233. QED. thanks