InterviewSolution
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InterviewSolution
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A consignment of 15 record players contains 4 defectives. The record players are selected at random, one by one, and examined. The one is examined, are not put back. Then the probability that 9th one examined is the last defective, isA. `(.^(4)C_(3)xx .^(11)C_(5))/(.^(15)C_(8))`B. `(.^(4)C_(3)xx .^(11)C_(5))/(.^(15)C_(8))xx(1)/(7)`C. `(.^(11)C_(5))/(.^(15)C_(8))xx(1)/(7)`D. `(.^(4)C_(3))/(.^(11)C_(5))xx(1)/(7)` |
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Answer» Correct Answer - B Let A and B be two events defined by A= Getting exactly 3 defectives in the examination of 8 record players. `B=9^(th)` record player is defective. Required probability `=P(A cap B)=P(A)P(B//A)` Now, `P(A)=(.^(4)C_(3)xx .^(11)C_(5))/(.^(15)C_(8))` and, `P(B//A)`=Probability that the 9th examined record player is defective given that there were 3 defective in the first 8 pieces examined `implies P(B//A)=(1)/(7)` `therefore` Required probability `=(.^(4)C_(3)xx .^(11)C_(5))/(.^(15)C_(8))xx(1)/(7)` |
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