Question
If $X \sim P\left(\frac{1}{2}\right)$, then find $P(X=3)$ given $e^{-0.5}=0.6065$.

Answer

$\begin{aligned} & \because \quad X \sim P\left(m=\frac{1}{2}\right) \\ & \therefore \quad p(x)=\frac{e^{-m} \cdot m^x}{x !} \\ & \because \quad P(X=3)=\frac{e^{-0.5} \times(0.5)^3}{3 !} \\ & =\frac{0.6065 \times 0.125}{=} \\ & =\frac{0.076}{6} \\ & =0.013 \\ & \end{aligned}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free