a shopkeeper received a pack of 15 pen out of which 4 were defective.the shopkeeper decided to examine every one one by one by selecting a pen at random .the pen examined are not put back.what is the probability that the ninth one examined is the last defective pen
options 11/195 b) 16/195 c)8/195 d)17/195
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Total number of cases is 15p9 cases ,
9 one is defective it can be selected and arranged in 4p1 ways , and similarly for the Remaining 8 places , there has to be 5 Non defective (11c5) and 3 defective (3c3) , they can be arranged in 8! Ways
(4P1*11C5*3c3*8!)/(15P9) =8/195
Answered by
1
Total number of cases is 15p9 cases ,
9 one is defective it can be selected and arranged in 4p1 ways , and similarly for the Remaining 8 places , there has to be 5 Non defective (11c5) and 3 defective (3c3) , they can be arranged in 8! Ways
(4P1*11C5*3c3*8!)/(15P9) =8/195
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