Question 485447
<pre>
Probability that all will be nondefective is

P(1st is nondefective)×P(2nd is nondefective)×P(3rd is nondefective)

(a) with replacement.  The probabilities do not change:

P(1st is nondefective)×P(2nd is nondefective)×P(3rd is nondefective) =

{{{expr(490/500)*expr(490/500)*expr(490/500)=
expr(49/50)*expr(49/50)*expr(49/50)=117649/125000=0.941192}}}
 
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(b) without replacement.  The probabilities do change:

P(1st is nondefective)×P(2nd is nondefective)×P(3rd is nondefective) =

{{{expr(490/500)*expr(489/499)*expr(488/498)=116929680/124251000=974414/1035425 =0.9410763696}}} 

So there is very little difference between the probabilies.

Edwin>/pre>