Q. A sample of a 4 times is drawn at a random without replacement from a lot of 10 items containing 3 defectives. If $x$ denotes the number of defective items in the sample, then $P(0 < x < 3)$is equal to
A) $\frac{3}{10}$
B) $\frac{4}{5}$
C) $\frac{1}{2}$
D) $\frac{1}{6}$
Explanation
(b)
$P(0 < x < 3) = P(x = 1) + P(x = 2)$
$= \frac{\ ^{3}C_{1}{\ \times}^{7}C_{3}}{\ ^{10}C_{4}} + \frac{\ ^{3}C_{2}{\ \times}^{7}C_{2}}{\ ^{10}C_{4}} = \frac{3 \times 35}{210} + \frac{3 \times 21}{210} = \frac{4}{5}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.