Question 118420
 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.
# of ways to pick the 3 cracked = 1
# of ways to pick 2 others = 9C2 = [9*8]/[1*2[ = 36
# of ways to pick 5 of 12 = 12C5
Therefore P(3 cracked and 2 not) = [1*36]/12C5 = 36/792 = 0.0455
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b. None of the cracked eggs are selected.
# of ways to pick 5 not cracked = 9C5
# of ways to pick 5 of 12 = 12C5
Prob(5 not cracked) = 9C5/12C5 = 0.1591
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c. Two of the cracked eggs are selected. 
# of ways to pick 2 cracked = 3C2= 3
# of ways to pick 3 not cracked = 9C3 = 84
# of ways to pick 5 of 12 = 12C5
Prob(2 cracked of 5 picked) = [3*84]/12C5 = 0.3182
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Cheers,
Stan H.