SOLUTION: A piece of equipment contains six identical items and it is known that three of them are defective. The items are tested one after the other until the three defective items are f

(a,i)  If the testing process is stopped on the third item, it means that the first item was defective, second item
       was defective and third item was defective. The probability of this event is


           P =  = .    ANSWER



(a,ii) If the testing process is stopped on the fourth item, it may happen in these three cases, shown by sequences
       of letters

               nDDD,  DnDD,  DDnD

       where "n" symbolizes non-defective item, "D" sympolises defective item.  So,


              P(process is stopped on the fourth item) = P(nDDD) + P(DnDD) + P(DDnD).


       Next,  P(nDDD) =  =  =  = ; 

              P(DnDD) =  = ;

              P(DDnD) =  =  = 


       Thus final probability   P(process is stopped on the fourth item) =  +  +  = .    ANSWER



(b)    If the testing process is stopped on the fourth item, it means that one of the three cases nDDD, DnDD, or DDnD happens,

       which has the probability of  ,  as it is shown in part  (a,ii)  above.


       If at the same time, the first item is not defective, it means that the event nDDD did not happens, but one of  DnDD or DDnD happens,

       which has the probability   +  = .

       THEREFORE, the  conditional probability  under given question is  P =  = .    ANSWER