A consignment of 15 record players contains 4 defectives. The record players are selected at random, one by one and examined. The one examined are not put back. Then

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Probability of getting exactly 3 defectives in the examination of 8 record players is $\frac{^{4} C_{3} \times^{11} C_{5}}{{}^{15}C_{8}}$

Probability $9^{t h}$ one examined is last defective is $\frac{8}{197}$

Probability that $9^{t h}$ examined record player is defective given that there were 3 defectives in the
First 8 players examined is $\frac{1}{7}$

Probability that $9^{t h}$ one examined is the last defective is $\frac{8}{195}$

answer is A, B, C.

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Detailed Solution

Let A be the event of getting exactly 3 defectives in the examination of 8 record players and B be the event $9^{t h}$ record player is defective
$P (A \cap B) = P (A) P (\frac{B}{A}) P (A) = \frac{^{4} C_{3} \times^{11} C_{5}}{^{15} C_{8}}, P (\frac{B}{A}) = \frac{1}{7}$