Q. A random variable $X$ has poisson distribution with mean 2. Then, $P(X > 1.5)$ equals
A) $\frac{3}{e^{2}}$
B) $1 - \frac{3}{e^{2}}$
C) 0
D) $\frac{2}{e^{2}}$
Explanation
(b)
$\therefore\ P(X > 1.5) = P(2) + P(3) + ...\infty$
$= 1 - \left\lbrack P(O) + P(1) \right\rbrack$
$= 1 - \left( e^{- 2} + \frac{e^{- 2} \times 2}{1} \right) = 1 - \frac{3}{e^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.