Practicing Success
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 |
$\frac{^4C_3× {^{11}C}_5}{^{15}C_8}$ $\frac{^4C_3× {^{11}C}_5}{^{15}C_8}×\frac{1}{7}$ $\frac{^{11}C_5}{^{15}C_8}×\frac{1}{7}$ $\frac{^{4}C_3}{^{11}C_5}×\frac{1}{7}$ |
$\frac{^4C_3× {^{11}C}_5}{^{15}C_8}×\frac{1}{7}$ |
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 ∴ Required probability $=\frac{^4C_3× {^{11}C}_5}{^{15}C_8}×\frac{1}{7}$ |