document.write( "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? \n" ); document.write( "
Algebra.Com's Answer #242954 by jrfrunner(365)\"\" \"About 
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You have 12 eggs of which 3 are cracked and 9 are not cracked.
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\n" ); document.write( "P(select 5 eggs of which 2 are cracked)=9C3*3C2/12C5
\n" ); document.write( "What this means
\n" ); document.write( "There are a total of 12C5 ways of selecting 5 eggs out of the 12 =792
\n" ); document.write( "Of the 5 eggs selected
\n" ); document.write( "There are 9C3 ways of selecting the 3 which are not cracked=84
\n" ); document.write( "And 3C2 of selecting the two eggs that are cracked=3
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\n" ); document.write( "So P (select 5 eggs of which 2 are cracked) =84*3/792=0.318
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\n" ); document.write( "\n" ); document.write( "Another way to look at this
\n" ); document.write( "Let C=event of selecting a cracked egg
\n" ); document.write( "Let N=event of selecting a non-cracked egg
\n" ); document.write( "You select 5 eggs and the probability of selecting 2 cracked eggs is
\n" ); document.write( "P(CCNNN) =3/12*2/11*9/10*8/9*7/8
\n" ); document.write( "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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\n" ); document.write( "Furthermore, the event of two broken eggs out of five can happen in many different ways
\n" ); document.write( "CCNNN
\n" ); document.write( "NNNCC
\n" ); document.write( "CNNNC
\n" ); document.write( "Etc
\n" ); document.write( "There are 5C2=10 ways of selecting the combination of 2 cracked eggs out of 5 eggs.
\n" ); document.write( "And each of those ways has the same probability as P(CCNNN)
\n" ); document.write( "Therefore
\n" ); document.write( "P(select 5 eggs of which 2 are cracked)=5C2*3/12*2/11*9/10*8/9*7/8 =0.318\r
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