A consignment of 15 record players contains 4 defectives. The record players are selected at random, one by one, examined. The ones examined are not put back. The probability 9th one examined is the last defective, is
Answer & explanation
Correct answer: option 2
Let A and B be two events defined by
A = Getting exactly 3 defectives in the examination of 8 record players.
B=9th record player is defective.
Required probability = $P(A∩ B)-P(A) P(B/A)$ ......(i)
Now,
$P(A) =\frac{^4C_3× {^{11}C}_5}{^{15}C_8}$
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
$⇒ P(B/A)=\frac{1}{7}$
∴ Required probability $=\frac{^4C_3× {^{11}C}_5}{^{15}C_8}×\frac{1}{7}$