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  • An urn contains 25 balls of which 10 balls bear a mark 'X' and the remaining 15 bear a mark 'Y' (iii) at least one ball will bear 'Y' mark.

Q. 5  An urn contains 25 balls of which 10 balls bear a mark  'X'  and the remaining 15 bear a mark 'Y'. A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that

             (iii) at least one ball will bear 'Y' mark.

Answers (1)

best_answer

 

P(atleast\, one\, bear\, \, Y\, mark) =P(Z\geq 1)=1-P(Z=0)

                                                                                             =1-(\frac{2}{5})^6

   

                  

Posted by

seema garhwal

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