A consignment of 15 record players contain 4 defectives. The record players are selected at random, one by one and examined. The one examined is not put back. Then : Find the Probability that `9^(th)` one examined is the last defective is `(8)/(195)`.

Options

A. `(.^(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)`

Answer

B

Explanation

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)`