Question 338902: a carton of eggs contains 12 eggs, 3 of which are cracked. what is the chance that when selecting five of these eggs for hard boiling, two of them are cracked?
Found 2 solutions by stanbon, jrfrunner:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! a carton of eggs contains 12 eggs, 3 of which are cracked. what is the chance that when selecting five of these eggs for hard boiling, two of them are cracked?
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P(a random egg being cracked) = 3/12 = 1/4
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Now, binomial problem with p = 1/4 and n = 5
P(x=2) = 5C2(1/4)^2(3/4)^3 = binompdf(5,1/4,2) = 0.2637
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Cheers,
Stan H.
Answer by jrfrunner(365)  (Show Source):
You can put this solution on YOUR website! You have 12 eggs of which 3 are cracked and 9 are not cracked.
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P(select 5 eggs of which 2 are cracked)=9C3*3C2/12C5
What this means
There are a total of 12C5 ways of selecting 5 eggs out of the 12 =792
Of the 5 eggs selected
There are 9C3 ways of selecting the 3 which are not cracked=84
And 3C2 of selecting the two eggs that are cracked=3
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So P (select 5 eggs of which 2 are cracked) =84*3/792=0.318
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Another way to look at this
Let C=event of selecting a cracked egg
Let N=event of selecting a non-cracked egg
You select 5 eggs and the probability of selecting 2 cracked eggs is
P(CCNNN) =3/12*2/11*9/10*8/9*7/8
because…you have to account for the fact that the chances of each choice changes due to non-replacement of the eggs.
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Furthermore, the event of two broken eggs out of five can happen in many different ways
CCNNN
NNNCC
CNNNC
Etc
There are 5C2=10 ways of selecting the combination of 2 cracked eggs out of 5 eggs.
And each of those ways has the same probability as P(CCNNN)
Therefore
P(select 5 eggs of which 2 are cracked)=5C2*3/12*2/11*9/10*8/9*7/8 =0.318
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