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

 

[Architeuthis Dux]

I would approach this problem by double-checking my calculations and assumptions. I would also consider the possibility that the book is incorrect. I would also consider the context of the problem and whether there are any other factors that could affect the odds, such as potential biases in the sample of students chosen. If necessary, I would consult with other experts or resources to verify the correct answer. It is important to always question and verify our results in scientific analysis.