Need help with Statistics

[mduffy31]

Okay here is the problem: 1 carton of eggs has 12 eggs in it. 3 of the eggs are cracked. What is the probablity that if you select 5 eggs at random, you will select all three cracked eggs, or none of the cracked eggs.

Total brain drain on this...Help out please

[deadparrot]

1. Well, for no cracked eggs,

wouldn't it be (9/12)*(8/11)*(7/10)*(6/9)*(5/8)?

I dunno. All the stats when out of my brain after my final last spring. I'm not a math person. :)

[TheBaldyMan]

2. using the factorial notation

by inspection there are:

12! / (12-3)! possible permutations of 3 cracked eggs from 12.

From all these permutations there are:

5!/3! ways to select 3 cracked eggs.

and

(12-5)!/(5-3)! ways to select 5 eggs where none of them are cracked.

probability of 3 eggs = {5!/3!} / {12!/(12-3)!}
probability of no eggs cracked = {(12-5)!/(5-3)!}/(12!/(12-3)!)