SOLUTION: I am having a hard time figuring out the solution to this problem.
A carton contains 12 eggs, 3 of which are cracked. If we randomly select 5 of the eggs for hard boiling, what
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A carton contains 12 eggs, 3 of which are cracked. If we randomly select 5 of the eggs for hard boiling, what
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Question 118420This question is from textbook
: I am having a hard time figuring out the solution to this problem.
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.
Any help is greatly appreciated!
This question is from textbook
You can put this solution on YOUR website! 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.
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