Question 1000809
{{{P(D)=3/33=1/11}}}
{{{P(D,D)=(1/11)(1/11)=1/121}}}
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{{{P(ND)=30/33=10/11}}}
Since you don't know whether the first part is defective or the second, you count both instances.
{{{P=P(D,ND)+P(ND,D)=(1/11)(10/11)+(10/11)(1/11)=20/121}}}
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The last remaining probability is neither part is defective.
{{{P(ND,ND)=(10/11)(10/11)=100/121}}}
If you sum all those probabilities, you will get 1, which shows you that these are the only three possible occurences.