Q. The mean and standard deviation of a binomial variate $X$ are 4 and $\sqrt{3}$ respectively. Then, $P(X \geq 1)$ is equal to
A) $1 - \left( \frac{1}{4} \right)^{16}$
B) $1 - \left( \frac{3}{4} \right)^{16}$
C) $1 - \left( \frac{2}{3} \right)^{16}$
D) $1 - \left( \frac{1}{3} \right)^{16}$
Explanation
(b)
Mean$= np = 4$, variance$= nqp = 3$
On solving, we get $q = \frac{3}{4},\ n = 16,p = \frac{1}{4}$
Now, $P(X \geq 1) = 1 - P(X = 0) = 1 -^{n}C_{0}p^{0}q^{n - 0}$
$= 1 - \left( \frac{3}{4} \right)^{16}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.