A carton contains 12 eggs, 3 of which are cracked.

[Timcago]

A carton contains 12 eggs, 3 of which are cracked. If we randomly select 5 of the eggs for hard boiling, what is the probability of the following events?

a. All of the cracked eggs are selected
b. None of the cracked eggs are selected.
c. Two of the cracked eggs are selected

I need this problem solved using either permutations or combinations because that is what we are learning. Can someone help?

 

[pka]

Using the standard notation for combinations, there are \(\displaystyle {{12} \choose 5}\) ways to select 5 from 12.

a. All of the cracked eggs are selected: \(\displaystyle {{3} \choose 3}{{9} \choose 2}\)
b. None of the cracked eggs are selected: \(\displaystyle {{9} \choose 5}\)
c. Two of the cracked eggs are selected: \(\displaystyle {{3} \choose 2}{{9} \choose 3}\)

 

[galactus]

#1: \(\displaystyle \L\\(\frac{3}{12})(\frac{2}{11})(\frac{1}{10})\underbrace{\frac{5!}{3!2!}}_{\text{ways to arrange\\chosen eggs}}\)


#2: \(\displaystyle \L\\(\frac{9}{10})(\frac{8}{11})(\frac{7}{10})(\frac{6}{9})(\frac{5}{8})\)


#3: \(\displaystyle \L\\(\frac{3}{12})(\frac{2}{11})(\frac{9}{10})(\frac{8}{9})(\frac{7}{8})\frac{5!}{3!2!}\)