A consignment of 15 record players contains 4 defectives. The record players are selected at random, one by one, and examined. The ones examined are not put back. The probability that 9th one examined is the last defective, is

# A consignment of $15$ record players contains $4$ defectives. The record players are selected at random, one by one, and examined. The ones examined are not put back. The probability that 9th one examined is the last defective, is

1. A

2. B

3. C

4. D

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### Solution:

Let $A$ and $B$ be two events defined by

A $=$ Getting exactly $3$ defectives in the examination of $8$ record players.

$=$ $9th$ record player is defective.

Required probability $=P\left(A\cap B\right)=P\left(A\right)P\left(B/A\right)$          …(i)

Now,

and,

$P\left(B/A\right)=$Probability that the $9th$ examined record player is defective given that there were $3$ defective in the first $8$ pieces examined

$\therefore$  Required probability

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