A consignment of 15 record players contains 4 defectives. The record players are selected at random one by one without replacement and examined

$\therefore$  Selection of 8 players in that 3 are defective is $n\left(E\right)={}^4{C}_3\times {}^{11}{C}_5$

$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{{}^4{C}_3\times {}^{11}{C}_5}{{}^{15}{C}_8}=\frac{8\times 7}{195}$

Now 9^th player can be selected who is defective is 1 way probability is $\frac{1}{7}$

$\therefore$  Total probability is $\frac{8\times 7}{195}\times \frac{1}{7}$

$=\frac{8}{195}$