Q. The probability that a certain kind of component will survive a given shock test is $\frac{3}{4}$. The probability that exactly 2 of the next 4 components tested survive is
A) $\frac{9}{41}$
B) $\frac{25}{128}$
C) $\frac{1}{5}$
D) $\frac{27}{128}$
Explanation
(d)
Required probability =$\ ^{4}C_{2}\left( \frac{3}{4} \right)^{2}\left( \frac{1}{4} \right)^{2}$
$= \frac{4!}{2!2!} \times \frac{9}{16} \times \frac{1}{16} = \frac{27}{128}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.